Find the sum of the series

Relevant wiki: Telescoping Series - Sum

Since $1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}$ the given infinite series has partial sums

$\sum_{n=1}^N \frac{2}{n(n+1)} = \sum_{n=1}^N \left(\frac{2}{n} - \frac{2}{n+1}\right) = \frac{2}{1} - \frac{2}{N+1} \xrightarrow{N\rightarrow\infty} 2.$

$\text{Let }a_n \text{ refer the n-th term in the above sequence of sum} \\ a_1 = 1 = 2 - \dfrac{2}{2} \\ a_2 = \dfrac{1}{3} = \dfrac{2}{2} - \dfrac{2}{3} \\ a_3 = \dfrac{1}{6} = \dfrac{2}{3} - \dfrac{2}{4} \\ \large \ldots \\ a_n = \dfrac{2}{n} - \dfrac{2}{n+1} \\ \text{Add up all those, till } n \rightarrow \infty \\ S = 2 - \dfrac{2}{2} + \dfrac{2}{2} - \dfrac{2}{3} + \dfrac{2}{3} - \dfrac{2}{4} + \ldots + \dfrac{2}{n} - \dfrac{2}{n+1} \\ S = 2 + \left(0 + 0 + 0 + \ldots + 0 + 0\right) ~ \boxed{\text{Since, }\dfrac{1}{\infty} = 0} \\ S = 2$

$\dfrac{1}{1}+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...\\ = \dfrac{1}{\frac{1 \cdot 2}{2}}+\dfrac{1}{\frac{2 \cdot 3}{2}}+\dfrac{1}{\frac{3 \cdot 4}{2}}+...\\ = 2(\dfrac{1}{1 \cdot 2}+\dfrac{1}{2 \cdot 3}+\dfrac{1}{3 \cdot 4}+...)\\ = 2(\dfrac{2-1}{1 \cdot 2}+\dfrac{3-2}{2 \cdot 3}+\dfrac{4-3}{3 \cdot 4}+...)\\ = 2(\dfrac{2}{1 \cdot 2}-\dfrac{1}{1 \cdot 2}+\dfrac{3}{2 \cdot 3}-\dfrac{2}{2 \cdot 3}+\dfrac{4}{3 \cdot 4}-\dfrac{3}{3 \cdot 4}+...)\\ = 2(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...)\\ = 2(\dfrac{1}{1})\\ = 2$

for k=500..10000 by 500
{
set sum to 0
for i=1..k by 1
{
set den to 0
set num to 1
for j=1..i by 1
{
change den by j
}
change sum by (num/den)
}
print sum
}

// observe convergence for large n to 2

In these problems, generalizing the series helps a lot.

denominator is nothing but sum to 'n' terms.

Then, we get, $\frac{2}{n(n+1)}$

When we have product of 2 terms in denominator, splitting them into 2 individual terms helps most of the time.

So, we get $\frac{2}{n}-\frac{2}{n+1}$ and n ranges from 1 to inf.

If we expand the series above, we get first term 2 and last term as (2/inf) and middle terms all cancels out.

We will get the final answer as $\boxed{2}$

1 + 2 + 3 + • • • + n = n( n + 1)/2 Simple, really.

My Java implementation:

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double j = 0;
double p = 0;
for (double i = 1; i < 2000; i++) {
    j += 1 / (p + i);
    p += i;
}
System.out.println(Math.round(j));