A nice problem to find a closed form for a rather challenging infinite series. Can you get it without using a computer?

Find the exact expression for the infinite sum \begin{equation} \frac{1}{1} + \frac{2}{2+3} + \frac{3}{4+5+6} + \frac{4}{7+8+9+10} + \dotsb \end{equation}

### Solution

The usual method starts by rewriting the infinite sum in a compact form, $$\sum_{n=1}^\infty f(n),$$ for an appropriate function $$f.$$ We do so after the following observations:

• The $$n$$-th term of the sum is a fraction, for which the numerator coincides with the number $$n.$$
• We can see the denominator of the $$n$$-th term as the difference of two sums. Notice that the largest term of those sums coincides with the sum of all the positive integers from one to $$n.$$ The smallest term in the sums of the denominators is one unit larger than the sum of all positive integers from one to $$n-1.$$ Analytically, this is the following expression:
\begin{equation} \underbrace{\dfrac{1}{2}\,\overbrace{\dfrac{n(n+1)}{2}}^{1+\dotsb+n}\bigg(\dfrac{n(n+1)}{2}+1\bigg)}_{1+\dotsb+\tfrac{n(n+1)}{2}}-\underbrace{\dfrac{1}{2}\,\overbrace{\dfrac{n(n-1)}{2}}^{1+\dotsb+(n-1)}\bigg(\dfrac{n(n-1)}{2}+1\bigg)}_{1+\dotsb+\frac{n(n-1)}{2}} \end{equation}

which reduces to $$\frac{1}{2}(n^3+n).$$

The last step is then to produce the sum of the expression

\begin{equation} \sum_{n=1}^\infty \frac{2n}{n^3+n} = \sum_{n=1}^\infty \frac{2}{n^2+1}. \end{equation}

The easy way is to obtain the aid of a mathematical software package (mathematica, maple, matlab, …)

For example, www.wolframalpha.com/input/?i=sum[2/(x^2%2B1),{x,1,infinity}] gives not only the answer ($$\pi \coth \pi -1$$), but also important information about the sum: Both the ratio and root tests are inconclusive, but the integral test indicates that the sum converges.

How would the reader use this information to obtain the value of the sum by traditional methods?  The integral test only offers a crude estimation for the sum:  Since we know that

\begin{equation} \int_1^\infty \frac{2}{x^2+1}\, dx \leq \sum_{n=1}^\infty \frac{2}{n^2+1} \leq 1 +\int_1^\infty \frac{2}{x^2+1}\, dx, \end{equation}

and in our case, as $$\int_1^\infty \tfrac{2}{x^2+1}\, dx =\frac{\pi}{2}$$, we have the estimate

\begin{equation} \frac{\pi}{2} \leq \sum_{n=1}^\infty \frac{2}{n^2+1} \leq 1+\frac{\pi}{2}. \end{equation}

We can go so much further with the Laplace transform: Notice first that the Laplace transform of $$\sin \omega t$$ is precisely the expression $$\omega (s^2+\omega^2)^{-1}.$$ In this case, we may use this fact to rewrite the infinite sum as follows:

\begin{equation} \sum_{n=1}^\infty \frac{2}{n^2+1} = 2 \sum_{n=1}^\infty \frac{1}{n^2+1} = 2 \sum_{n=1}^\infty \int_{0}^\infty e^{-nx} \sin x\, dx \end{equation}

Since we have proved that the series converges, we are allowed to change integral with summation. We obtain

\begin{equation} \sum_{n=1}^\infty \frac{2}{n^2+1} = 2 \int_0^\infty \sin x \sum_{n=1}^\infty e^{-nx}\, dx = 2\int_0^\infty \frac{e^{-x}}{1-e^{-x}}\,\sin x \, dx \end{equation}

The key is therefore to find the last integral in the expression above.