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## How do I find the sum of sum_4^(oo)(1/(k-3)-1/(k))?

Steve M
Featured 1 month ago

${\sum}_{4}^{n} \left(\frac{1}{k - 3} - \frac{1}{k}\right) = \frac{\left(n - 3\right) \left(11 {n}^{2} - 18 n + 4\right)}{6 n \left(n - 1\right) \left(n - 2\right)}$

${\sum}_{4}^{\infty} \left(\frac{1}{k - 3} - \frac{1}{k}\right) = \frac{11}{6}$

#### Explanation:

We seek:

$S = {\sum}_{4}^{\infty} \left(\frac{1}{k - 3} - \frac{1}{k}\right)$

Firstly, let us find a formula for:

$S \left(n\right) = {\sum}_{4}^{n} \left(\frac{1}{k - 3} - \frac{1}{k}\right)$

This a a typical summation of differences, and as such many terms will cancel. If we expand the summation this becomes clear:

$S \left(n\right) = \left(\frac{1}{1} - \textcolor{b l u e}{\frac{1}{4}}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = 4\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\frac{1}{2} - \textcolor{g r e e n}{\frac{1}{5}}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = 5\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\frac{1}{3} - \textcolor{red}{\frac{1}{6}}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = 6\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\textcolor{b l u e}{\frac{1}{4}} - \frac{1}{7}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = 7\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\textcolor{g r e e n}{\frac{1}{5}} - \frac{1}{8}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = 8\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\textcolor{red}{\frac{1}{6}} - \frac{1}{9}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = 9\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \vdots$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\frac{1}{n - 6} - \textcolor{p u r p \le}{\frac{1}{n - 3}}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \left(k = n - 3\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\frac{1}{n - 5} - \frac{1}{n - 2}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \left(k = n - 2\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\frac{1}{n - 4} - \frac{1}{n - 1}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = n - 1\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\textcolor{p u r p \le}{\frac{1}{n - 3}} - \frac{1}{n}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = n\right)$

And now if we cancel the terms which appear as positive from one term and negative and another:

$S \left(n\right) = \left(\frac{1}{1} - \textcolor{b l u e}{\cancel{\frac{1}{4}}}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = 4\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\frac{1}{2} - \textcolor{g r e e n}{\cancel{\frac{1}{5}}}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = 5\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\frac{1}{3} - \textcolor{red}{\cancel{\frac{1}{6}}}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = 6\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\textcolor{b l u e}{\cancel{\frac{1}{4}}} - \cancel{\frac{1}{7}}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = 7\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\textcolor{g r e e n}{\cancel{\frac{1}{5}}} - \cancel{\frac{1}{8}}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = 8\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\textcolor{red}{\cancel{\frac{1}{6}}} - \cancel{\frac{1}{9}}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = 9\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \vdots$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\cancel{\frac{1}{n - 6}} - \textcolor{p u r p \le}{\cancel{\frac{1}{n - 3}}}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \left(k = n - 3\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\cancel{\frac{1}{n - 5}} - \frac{1}{n - 2}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \left(k = n - 2\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\cancel{\frac{1}{n - 4}} - \frac{1}{n - 1}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = n - 1\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\textcolor{p u r p \le}{\cancel{\frac{1}{n - 3}}} - \frac{1}{n}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = n\right)$

Which leaves us with a much reduced collection of terms:

$S \left(n\right) = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} - \frac{1}{n - 2} - \frac{1}{n - 1} - \frac{1}{n}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{11}{6} - \frac{1}{n - 2} - \frac{1}{n - 1} - \frac{1}{n}$ ..... [A]

Although not strictly required, We can simplify this result (this is typically the case in an exam question when $S \left(n\right)$ is sought prior to finding the limiting case $S \left(\infty\right)$)

$\setminus \setminus = \frac{11}{6} - \frac{1}{n - 2} - \frac{1}{n - 1} - \frac{1}{n}$

$\setminus \setminus = \frac{11 n \left(n - 1\right) \left(n - 2\right) - 6 n \left(n - 1\right) - 6 n \left(n - 2\right) - 6 \left(n - 1\right) \left(n - 2\right)}{6 n \left(n - 1\right) \left(n - 2\right)}$

$\setminus \setminus = \frac{11 n \left(n - 1\right) \left(n - 2\right) - 6 n \left(n - 1\right) - 6 n \left(n - 2\right) - 6 \left(n - 1\right) \left(n - 2\right)}{6 n \left(n - 1\right) \left(n - 2\right)}$

$\setminus \setminus = \frac{11 {n}^{3} - 33 {n}^{2} + 22 n - 6 {n}^{2} + 6 n - 6 {n}^{2} + 12 n - 6 {n}^{2} + 18 n - 12}{6 n \left(n - 1\right) \left(n - 2\right)}$

$\setminus \setminus = \frac{11 {n}^{3} - 51 {n}^{2} + 58 n - 12}{6 n \left(n - 1\right) \left(n - 2\right)}$

$\setminus \setminus = \frac{\left(n - 3\right) \left(11 {n}^{2} - 18 n + 4\right)}{6 n \left(n - 1\right) \left(n - 2\right)}$

Factorization was performed via a calculator at the last step, the answer at [A] would suffice

So finally, we seek:

$S = {\lim}_{n \rightarrow \infty} S \left(n\right)$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \left\{\frac{11}{6} - \frac{1}{n - 2} - \frac{1}{n - 1} - \frac{1}{n}\right\} \setminus \setminus \setminus \setminus \setminus \setminus \setminus$ (from [A])
$\setminus \setminus = \frac{11}{6}$

||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Validation:

It is always worth checking a few sums just to check the result is consistent

By direct summation
S(4) = 3/4
S(5) = 21/21
S(5) = 73/60

And using the derived summation formula gives the same result

## Find gradient of the curve ?y^3+y=x^3+x^2 at the point where the curve meets the coordinate axes

Steve M
Featured 1 month ago

At coordinate $\left(0 , 0\right)$ we have $y = 0$

At coordinate $\left(- 1 , 0\right)$ we have $y = x + 1$

#### Explanation:

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point.

We have a curve, $C$ say, given by:

$C : \setminus \setminus \setminus {y}^{3} + y = {x}^{3} + {x}^{2}$

$C$ will meet the $x$-axis when $y = 0$

$\implies {x}^{3} + {x}^{2} = 0$
$\therefore {x}^{2} \left(x + 1\right) = 0$
$\therefore x = 0 , - 1$

$C$ will meet the $y$-axis when $x = 0$

$\implies {y}^{3} + y = 0$
$\therefore y \left({y}^{2} + 1\right) = 0$
$\therefore y = 0 \setminus \setminus \setminus$ (for $y \in \mathbb{R}$).

So, we have two coordinates $\left(0 , 0\right)$ and $\left(- 1 , 0\right)$

Next, we differentiate (implicitly) the given equation for $C$

$3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} + 2 x$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} \left(3 {y}^{2} + 1\right) = 3 {x}^{2} + 2 x$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {x}^{2} + 2 x}{3 {y}^{2} + 1}$

For both coordinates we will use the point/slope form $y - {y}_{1} = m \left(x - {x}_{1}\right)$.

At coordinate $\left(0 , 0\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{0}{0 + 1} = 0$

The tangent equation is therefore:

$y - 0 = 0 \left(x - 0\right)$
$y = 0$

At coordinate $\left(- 1 , 0\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 - 2}{0 + 1} = 1$

The tangent equation is therefore:

$y - 0 = 1 \left(x - \left(- 1\right)\right)$
$y = x + 1$

## How do you find the second derivative of s=((t+1)/(t+2))^2?

Michael F.
Featured 1 month ago

The second derivative is $s ' ' \left(t\right) = \frac{- 4 t - 2}{{\left(t + 2\right)}^{4}}$.

#### Explanation:

Let $s \left(t\right) = {\left(\frac{t + 1}{t + 2}\right)}^{2} = \frac{{\left(t + 1\right)}^{2}}{{\left(t + 2\right)}^{2}}$.

For differentiating we use the $\textcolor{g r e e n}{\text{chain rule}}$

color(green)((u(v(x))^' = u^'(v(x)) * v^'(x)  („outer“ times „inner“ derivative)

and the $\textcolor{b l u e}{\text{quotient rule}}$

$\textcolor{b l u e}{{\left(\frac{u \left(x\right)}{v \left(x\right)}\right)}^{'} = \frac{{u}^{'} \left(x\right) \cdot v \left(x\right) - u \left(x\right) \cdot {v}^{'} \left(x\right)}{v \left(x\right)} ^ 2}$ (with $v \left(x\right) \ne 0$).

The first derivative is calculated by

$s ' \left(t\right) = \frac{2 \cdot \left(t + 1\right) \cdot 1 \cdot {\left(t + 2\right)}^{2} - {\left(t + 1\right)}^{2} \cdot 2 \cdot \left(t + 2\right) \cdot 1}{{\left(t + 2\right)}^{4}}$

$s ' \left(t\right) = \frac{2 \cdot \left(t + 1\right) \cdot {\left(t + 2\right)}^{2} - {\left(t + 1\right)}^{2} \cdot 2 \cdot \left(t + 2\right)}{{\left(t + 2\right)}^{4}}$

Factorizing $\left(t + 2\right)$ from both summands of the enumerator gives us

$s ' \left(t\right) = \frac{\left(t + 2\right) \cdot \left[2 \cdot \left(t + 1\right) \cdot \left(t + 2\right) - {\left(t + 1\right)}^{2} \cdot 2\right]}{{\left(t + 2\right)}^{4}}$

Cancelling gives us

$s ' \left(t\right) = \frac{2 \cdot \left({t}^{2} + 3 t + 2\right) - 2 \cdot \left({t}^{2} + 2 t + 1\right)}{{\left(t + 2\right)}^{3}}$

$s ' \left(t\right) = \frac{\cancel{2 {t}^{2}} + 6 t + 4 \cancel{- 2 {t}^{2}} - 4 t - 2}{{\left(t + 2\right)}^{3}}$

$s ' \left(t\right) = \frac{2 t + 2}{{\left(t + 2\right)}^{3}}$.

The second derivative is calculated by

$s ' ' \left(t\right) = \frac{2 \cdot {\left(t + 2\right)}^{3} - \left(2 t + 2\right) \cdot 3 \cdot {\left(t + 2\right)}^{2} \cdot 1}{{\left(t + 2\right)}^{6}}$.

Factorizing ${\left(t + 2\right)}^{2}$ from both summands of the enumerator gives us

$s ' ' \left(t\right) = \frac{{\left(t + 2\right)}^{2} \cdot \left[2 \cdot \left(t + 2\right) - 3 \cdot \left(2 t + 2\right)\right]}{{\left(t + 2\right)}^{6}}$.

$s ' ' \left(t\right) = \frac{2 \cdot \left(t + 2\right) - 3 \cdot \left(2 t + 2\right)}{{\left(t + 2\right)}^{4}}$

$s ' ' \left(t\right) = \frac{2 t + 4 - 6 t - 6}{{\left(t + 2\right)}^{4}}$.

$\textcolor{red}{s ' ' \left(t\right) = \frac{- 4 t - 2}{{\left(t + 2\right)}^{4}}}$.

I hope it helps.

## What is the equation of the line normal to  f(x)=1/(e^x+4) at  x=6?

ASAP
Featured 2 weeks ago

$y = \left({\left({e}^{6} + 4\right)}^{2} / {e}^{6}\right) \left(x - 6\right) + \left(\frac{1}{{e}^{6} + 4}\right)$

#### Explanation:

To find the equation of Normal to the line f(x)=1/(e^x+4 we need to find the slope of the normal and the $y$ coordinate at ${x}_{1} = 6$

We know the slope of normal is $\frac{- 1}{\frac{\mathrm{dy}}{\mathrm{dx}}}$ so ,

$y = \frac{1}{{e}^{x} + 4}$

Differentiate both sides with respect to $x$ using the chain rule.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- {e}^{x}}{{e}^{x} + 4} ^ 2$

So the slope of the normal ( ${m}_{n}$) is $\frac{- 1}{\frac{\mathrm{dy}}{\mathrm{dx}}} = \frac{- 1}{\frac{- {e}^{x}}{{e}^{x} + 4} ^ 2}$

Therefore ${m}_{n} = {\left({e}^{x} + 4\right)}^{2} / {e}^{x}$

Now to find the ${y}_{1}$ coordinate when ${x}_{1} = 6$

We know $f \left(x\right) = \frac{1}{{e}^{x} + 4}$

$f \left(6\right) = \frac{1}{{e}^{6} + 4}$

Therefore the ${y}_{1}$ coordinate is $\frac{1}{{e}^{6} + 4}$

We know the equation of a line is

$\left(y - {y}_{1}\right) = {m}_{n} \left(x - {x}_{1}\right)$

$\left(y - \frac{1}{{e}^{6} + 4}\right) = \left({\left({e}^{6} + 4\right)}^{2} / {e}^{6}\right) \left(x - 6\right)$

Therefore $y = \left({\left({e}^{6} + 4\right)}^{2} / {e}^{6}\right) \left(x - 6\right) + \left(\frac{1}{{e}^{6} + 4}\right)$

## How do you determine if the series the converges conditionally, absolutely or diverges given Sigma (-1)^(n+1)/(n+1)^2 from [1,oo)?

Cesareo R.
Featured 2 weeks ago

See below.

#### Explanation:

Considering that

$L = {\sum}_{k = 1}^{\infty} \frac{1}{k} ^ 2 = {\pi}^{2} / 6$ (Basel problem)

https://en.wikipedia.org/wiki/Basel_problem

we have

$L = {\sum}_{k = 1}^{\infty} \frac{1}{2 k} ^ 2 + {\sum}_{k = 1}^{\infty} \frac{1}{2 k - 1} ^ 2 = {S}_{p} + {S}_{i}$

but

${S}_{p} = \frac{1}{4} L$ then

${S}_{p} - {S}_{i} = \frac{1}{4} L - \left(L - \frac{1}{4} L\right) = - \frac{L}{2} = - {\pi}^{2} / 12$

then

${\sum}_{k = 1}^{\infty} {\left(- 1\right)}^{k} / {k}^{2} = - {\pi}^{2} / 12$ so converges

## How do you find the intervals of increasing and decreasing using the first derivative given y=(x+2)^2(x-1)?

Gustaaf Vocking
Featured 2 weeks ago

$y ' > 0 \text{ for } x < - 2 \mathmr{and} x > 0$
$y ' < 0 \text{ for } x \succ 2 \mathmr{and} x < 0$

#### Explanation:

$y = {\left(x + 2\right)}^{2} \left(x - 1\right)$

A function is increasing where its derivative is positive, decreasing where its derivative is negative and has a (local or global) minimum or maximum when its derivative is 0.

Remember the product rule for differentiation:
$h \left(x\right) = f \left(x\right) \cdot g \left(x\right) \implies h ' \left(x\right) = f ' \left(x\right) \cdot g \left(x\right) + f \left(x\right) \cdot g ' \left(x\right)$

Let's define:
$f \left(x\right) = {\left(x + 2\right)}^{2} = {x}^{2} + 4 x + 4 \mathmr{and} g \left(x\right) = x - 1$
With:
$f ' \left(x\right) = 2 x + 4 \mathmr{and} g ' \left(x\right) = 1$

Then:
$y ' = \left(2 x + 4\right) \cdot \left(x - 1\right) + {\left(x + 2\right)}^{2} \cdot 1$
$= 2 {x}^{2} - 2 x + 4 x - 4 + {x}^{2} + 4 x + 4$
$= 3 {x}^{2} + 6 x$

Finding the (local or global) minima/maxima:
$y ' = 0 \text{ for } x = 0 \mathmr{and} x = - 2$
We then plug in values to find that:
$y ' > 0 \text{ for } x < - 2 \mathmr{and} x > 0$
$y ' < 0 \text{ for } x \succ 2 \mathmr{and} x < 0$

This implies that $x = - 2 \implies y = 0$ is a local maximum, and that $x = 0 \implies y = - 4$ is a local minimum.

In graphs:
$y ' :$
graph{3x^2 + 6x [-5, 5, -3.88, 1.12]}

$y :$
graph{ (x+2)^2 (x-1) [-10, 10, -6.38, 3.62]}

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