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Featured 1 month ago

# sum_4^(n)(1/(k-3)-1/(k)) = ( (n-3)(11n^2-18n+4) ) / ( 6n(n-1)(n-2) )#

# sum_4^(oo)(1/(k-3)-1/(k)) = 11/6 #

We seek:

# S = sum_4^(oo)(1/(k-3)-1/(k)) #

Firstly, let us find a formula for:

# S(n) = sum_4^(n)(1/(k-3)-1/(k)) #

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

# S(n) = (1/1 - color(blue)(1/4)) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=4) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ (1/2 - color(green)(1/5)) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=5) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ (1/3 - color(red)(1/6)) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=6) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ (color(blue)(1/4) - 1/7) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=7) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ (color(green)(1/5) - 1/8) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=8) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ (color(red)(1/6) - 1/9) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=9) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ vdots #

# \ \ \ \ \ \ \ \ \ \ \ \ \ (1/(n-6)-color(purple)(1/(n-3))) + \ \ \ \ \ \ (k=n-3) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ (1/(n-5)-1/(n-2)) + \ \ \ \ \ \ (k=n-2) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ (1/(n-4)-1/(n-1)) + \ \ \ \\ \ \ (k=n-1) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ (color(purple)(1/(n-3))-1/(n)) + \ \ \ \ \ \ \ \ \ \ \ \ \ (k=n) #

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

# S(n) = (1/1 - color(blue)(cancel(1/4))) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=4) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ (1/2 - color(green)(cancel(1/5))) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=5) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ (1/3 - color(red)(cancel(1/6))) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=6) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ (color(blue)(cancel(1/4)) - cancel(1/7)) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=7) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ (color(green)(cancel(1/5)) - cancel(1/8)) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=8) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ (color(red)(cancel(1/6)) - cancel(1/9)) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=9) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ vdots #

# \ \ \ \ \ \ \ \ \ \ \ \ \ (cancel(1/(n-6))-color(purple)(cancel(1/(n-3)))) + \ \ \ \ \ \ (k=n-3) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ (cancel(1/(n-5))-1/(n-2)) + \ \ \ \ \ \ (k=n-2) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ (cancel(1/(n-4))-1/(n-1)) + \ \ \ \\ \ \ (k=n-1) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ (color(purple)(cancel(1/(n-3)))-1/(n)) + \ \ \ \ \ \ \ \ \ \ \ \ \ (k=n) #

Which leaves us with a much reduced collection of terms:

# S(n) = 1/1 + 1/2 +1/3 -1/(n-2) -1/(n-1) -1/n #

# \ \ \ \ \ \ \ \ = 11/6 -1/(n-2) -1/(n-1) -1/n # ..... [A]

Although not strictly required, We can simplify this result (this is typically the case in an exam question when

# \ \ = 11/6 -1/(n-2) -1/(n-1) -1/n #

# \ \ = ( 11n(n-1)(n-2) - 6n(n-1) -6n(n-2) - 6(n-1)(n-2) ) / ( 6n(n-1)(n-2) )#

# \ \ = ( 11n(n-1)(n-2) - 6n(n-1) -6n(n-2) - 6(n-1)(n-2) ) / ( 6n(n-1)(n-2) )#

# \ \ = ( 11n^3-33n^2+22n -6n^2+6n -6n^2+12n -6n^2+18n-12 ) / ( 6n(n-1)(n-2) )#

# \ \ = ( 11n^3-51n^2+58n-12 ) / ( 6n(n-1)(n-2) )#

# \ \ = ( (n-3)(11n^2-18n+4) ) / ( 6n(n-1)(n-2) )#

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

So finally, we seek:

# S = lim_(n rarr oo) S(n) #

# \ \ = lim_(n rarr oo) {11/6 -1/(n-2) -1/(n-1) -1/n }\ \ \ \ \ \ \ # (from [A])

# \ \ = 11/6#

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**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

Featured 1 month ago

At coordinate

At coordinate

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: \ \ \ y^3+y=x^3+x^2 #

# => x^3+x^2 = 0 #

# :. x^2(x+1)= 0 #

# :. x = 0, -1 #

# => y^3+y = 0 #

# :. y(y^2+1)= 0 #

# :. y = 0 \ \ \ # (for#y in RR# ).

So, we have two coordinates

Next, we differentiate (implicitly) the given equation for

# 3y^2dy/dx+dy/dx = 3x^2+2x #

# :. dy/dx(3y^2+1) = 3x^2+2x #

# :. dy/dx = (3x^2+2x)/(3y^2+1) #

For both coordinates we will use the point/slope form

**At coordinate**

# dy/dx = (0)/(0+1) = 0#

The tangent equation is therefore:

# y- 0 = 0(x-0) #

# y = 0 #

**At coordinate**

# dy/dx = (3-2)/(0+1) = 1#

The tangent equation is therefore:

# y- 0 = 1(x-(-1)) #

# y = x+1 #

Featured 1 month ago

The second derivative is

Let

For differentiating we use the

and the

The first derivative is calculated by

Factorizing

Cancelling gives us

The second derivative is calculated by

Factorizing

Cancelling leads to

We receive the final result

I hope it helps.

Featured 2 weeks ago

To find the equation of Normal to the line

We know the slope of normal is

Differentiate both sides with respect to

So the slope of the normal (

Therefore

Now to find the

We know

Therefore the

We know the equation of a line is

Therefore

Featured 2 weeks ago

See below.

Considering that

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

we have

but

then

Featured 2 weeks ago

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:

Let's define:

With:

Then:

Finding the (local or global) minima/maxima:

We then plug in values to find that:

This implies that

In graphs:

graph{3x^2 + 6x [-5, 5, -3.88, 1.12]}

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

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