# Calculate $\sum\limits_{n\geq 1}\frac{1}{1+2+\ldots+n}$

Calculate $\displaystyle \sum_{n\geq 1}\dfrac{1}{1+2+\ldots+n}$

My attempts :

\begin{aligned} \sum_{n\geq 1}\dfrac{1}{1+2+\ldots+n} &= \sum_{n\geq 1}\dfrac{2}{n(n+1)}=2\sum_{n\geq 1}\left( \dfrac{1}{n}-\dfrac{1}{n+1}\right)=2\left( \dfrac{1}{1}-\dfrac{1}{+\infty}\right)=2(1-0)=2? \end{aligned}

• Both $\infty$ :D – Cauchy Jul 10 '17 at 11:31
• Try computing a few values of $$\sum_{n=1}^N\left(\frac1n-\frac1{n+1}\right)$$ just for kicks... – Did Jul 10 '17 at 11:38
• yes i see now thanks – Yacob Jul 10 '17 at 12:18

\begin{aligned} \sum_{n\geq 1}\dfrac{2}{n(n+1)}=2\sum_{n\geq 1}\dfrac{1}{n}-2\sum_{n\geq 1}\dfrac{1}{n+1} \end{aligned}

False. The equality does not hold because the two sums on the right are divergent, while the sum on the left is convergent.

Better go at it this way:

$$\sum_{n=1}^\infty\frac{1}{n(n+1)} = \sum_{n=1}^\infty\frac{1}{n}-\frac{1}{n+1} = \left(\frac 11 - \frac12\right) + \left(\frac12-\frac13\right) + \left(\frac13-\frac14\right) + \cdots$$

Now, remember that $$\sum_{n=1}^\infty a_n = \lim_{N\to\infty} \sum_{n=1}^N a_n$$

In your case, $$\sum_{n=1}^N \frac 1n-\frac{1}{n+1}$$ can be calculated relatively easily.

The series 'telescopes' with the only remaining terms being $1-\frac{1}{r+1}$ where $r$ is the number of terms. That's $\frac{r}{r+1}$ which tends to $1$ for partial sums. It's not too hard to show that this sum is not commutative because it alternates with decreasing terms.

$$\sum_{n>1}\frac{1}{1+2+...+n}=2\sum_{n>1}\left(\frac{1}{n}-\frac{1}{n+1}\right)=2\lim_{n\rightarrow+\infty}\left(1-\frac{1}{n+1}\right)=2$$

I like to look at it like this:

$$\sum_{n=1}^\infty\frac1{n(n+1)}=\sum_{n=1}^\infty\left(\frac1n-\frac1{n+1}\right)$$

\begin{align}\require{cancel}&=\left(\frac11-\frac12\right)+\left(\frac12-\frac13\right)+\left(\frac13-\frac14\right)+\dots\\&=\left(\frac11-\color{red}{\cancel{\frac12}}\right)+\left(\color{red}{\cancel{\frac12}}-\color{orange}{\cancel{\frac13}}\right)+\left(\color{orange}{\cancel{\frac13}}-\color{green}{\cancel{\frac14}}\right)+\color{green}\dots\\&=1\end{align}

Notably, this is called a telescoping series.

HINT:

$$\dfrac1{n(n+1)}=\dfrac{n+1-n}{n(n+1)}=\dfrac1n-\dfrac1{n+1}$$

• i already did that – Yacob Jul 10 '17 at 11:33
• @Yacob, Set a few values of $n$ to find the surviving terms – lab bhattacharjee Jul 10 '17 at 11:34
• – lab bhattacharjee Jul 10 '17 at 11:35