# How do you prove that: 1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...+n) = (2n)/(n+1) ?

Sep 26, 2016

True

#### Explanation:

Supposing that the afirmation is true

$1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \frac{1}{1 + 2 + 3 + 4} + \cdots + \frac{1}{1 + 2 + 3 + 4 + \cdots + n} = {\sum}_{k = 1}^{n} \frac{2}{k \left(k + 1\right)} = {S}_{n}$

${S}_{1} = \frac{2 \times 1}{1 + 1} = 1$

so supposing is true

${S}_{n} = \frac{2 n}{n + 1}$ then

${S}_{n + 1} = {S}_{n} + \frac{2}{\left(n + 1\right) \left(n + 2\right)} = \frac{2 \left(n + 1\right)}{n + 2}$ which is the expected result.

Then by mathematical induction, it is true the assertion.

Sep 26, 2016

Prove by induction.

#### Explanation:

Proof by induction:

Base case

If $n = 1$ then then left hand side is:

$1$

and the right hand side is:

$\frac{2 \left(1\right)}{\left(1\right) + 1} = \frac{2}{2} = 1$

So the equation holds for $n = 1$

Induction step

First note that:

$1 + 2 + 3 + \ldots + n = \frac{1}{2} n \left(n + 1\right)$

Suppose:

$1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \ldots + \frac{1}{1 + 2 + 3 + \ldots + n} = \frac{2 n}{n + 1}$

Then:

$1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \ldots + \frac{1}{1 + 2 + 3 + \ldots + n + \left(n + 1\right)}$

$= \frac{2 n}{n + 1} + \frac{1}{1 + 2 + 3 + \ldots + n + \left(n + 1\right)}$

$= \frac{2 n}{n + 1} + \frac{1}{\frac{1}{2} \left(n + 1\right) \left(n + 2\right)}$

$= \frac{2 n}{n + 1} + \frac{2}{\left(n + 1\right) \left(n + 2\right)}$

$= \frac{\left(2 n\right) \left(n + 2\right)}{\left(n + 1\right) \left(n + 2\right)} + \frac{2}{\left(n + 1\right) \left(n + 2\right)}$

$= \frac{2 {n}^{2} + 4 n + 2}{\left(n + 1\right) \left(n + 2\right)}$

$= \frac{2 \left({n}^{2} + 2 n + 1\right)}{\left(n + 1\right) \left(n + 2\right)}$

$= \frac{2 {\left(n + 1\right)}^{2}}{\left(n + 1\right) \left(n + 2\right)}$

$= \frac{2 \left(n + 1\right)}{n + 2}$

Sep 26, 2016

$\text{ The Sum=} \frac{2 n}{n + 1} .$

#### Explanation:

We will use a Method called Method of Telescopic Sum.

Denoting by, ${T}_{j}$, the ${j}^{t h}$ term of the Series, we observe that,

T_j=1/(1+2+...+j)=1/(sumj)=1/(j/2(j+1))=2/(j(j+1).

$\therefore \text{ The Reqd. Sum, say S} = {\sum}_{j = 1}^{j = n} {T}_{j} = {\sum}_{j = 1}^{j = n} \frac{2}{j \left(j + 1\right)}$

$= 2 {\sum}_{j = 1}^{j = n} \frac{1}{j \left(j + 1\right)} = 2 {\sum}_{j = 1}^{j = n} \frac{\left(j + 1\right) - j}{j \left(j + 1\right)}$

$= 2 {\sum}_{j = 1}^{j = n} \left\{\frac{j + 1}{j \left(j + 1\right)} - \frac{j}{j \left(j + 1\right)}\right\}$

$= 2 {\sum}_{j = 1}^{j = n} \left\{\frac{1}{j} - \frac{1}{j + 1}\right\}$

$= 2 \left[\left(\frac{1}{1} - \cancel{\frac{1}{2}}\right) + \left(\cancel{\frac{1}{2}} - \cancel{\frac{1}{3}}\right) + \left(\cancel{\frac{1}{3}} - \cancel{\frac{1}{4}}\right) + \ldots + \left(\cancel{\frac{1}{n - 1}} - \cancel{\frac{1}{n}}\right) + \left(\cancel{\frac{1}{n}} - \frac{1}{n + 1}\right)\right]$

$= 2 \left(1 - \frac{1}{n + 1}\right)$

$\therefore \text{ The Sum=} \frac{2 n}{n + 1}$, as Respected George C., and,

Respected Cesareo R. , have derived!

Enjoy Maths.!