Which (convergent) series can one find the sum of?

I know about geometric series and how one can find the sum when they are convergent.

I also have heard that one can prove that the $p$-series $$\sum_{n=1}^{\infty} \frac{1}{n^2}$$ has sum $\pi^2/6$ (but I don't know how this is actually proved).

This might be an unanswerable question, but are there any tools or general rules that will tell you when you can find the exact sum of a given series? Are there, for example, rules/tools for how to find the exact value of any $p$-series? (like with geometric series).

• Google "Riemann zeta function" – GFauxPas Apr 22 '15 at 12:55
• Consider the Fourier-Series of $f(x):=(x-\pi)^2$ defined on $[0,2\pi]$ and evaluate in zero or google for "Basler Problem" for the case $p=2$. – frog Apr 22 '15 at 12:55
• No great general rules, no. That would be too easy. – Thomas Andrews Apr 22 '15 at 12:55
• I hope you are aware of general concepts of convergence of an infinite series! – Jesse P Francis Apr 22 '15 at 13:01
• I agree with @ThomasAndrews, no general rule. The most likely cases (arguably easiest cases) are those when you can recognize the series as the Taylor series of some function evaluated at a point. For example $\sum_{n=0}^{\infty}\frac{1}{n!}=e^{1}$ because $\sum_{n=0}^{\infty}\frac{x^{n}}{n!}=e^{x}$. – TravisJ Apr 22 '15 at 13:09

Thomas Andrews is right when he says that there are no general rules. Often this type of problem, we must use a more intuitive method.

I know the p-series for p = 2. The resolution method in this series comes from Euler himself.

(1) $\sin(x) = x - x^3/3! + x^5/5! -x^7/7! + ...$ (Taylor series)

(2) $\sin(x)/x = 1 - x^2/3! + x^4/5! -x^6/7! + ...$ (divided by x)

(3) Now, the roots (intersections with the x-axis) of sin(x)/x occur precisely at x = $n\pi$ where n = ±1,±2,±3,... Let us assume we can express this infinite series as a (normalized) product of linear factors given by its roots, just as we do for finite polynomials

\begin{align} \frac{\sin(x)}{x} & {} = \left(1 - \frac{x}{\pi}\right)\left(1 + \frac{x}{\pi}\right)\left(1 - \frac{x}{2\pi}\right)\left(1 + \frac{x}{2\pi}\right)\left(1 - \frac{x}{3\pi}\right)\left(1 + \frac{x}{3\pi}\right) \cdots \\ & {} = \left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{4\pi^2}\right)\left(1 - \frac{x^2}{9\pi^2}\right) \cdots. \end{align}

(4) If we formally multiply out this product and collect all the $x^2$ terms (we are allowed to do so because of Newton's identities), we see that the $x^2$ coefficient of sin(x)/x is

$$1/\pi^2 + 1/(4\pi^2) + 1/(9\pi^2) + ... = (1/\pi^2)\sum_{n=1}^\infty\frac{1}{n^2}$$

(5) But from the original infinite series expansion of sin(x)/x, the coefficient of $x^2$ is −1/(3!) = −1/6. These two coefficients must be equal...

$$-1/6 = -(1/\pi^2)\sum_{n=1}^\infty\frac{1}{n^2}$$

(6) Multiplying through both sides of this equation by $-\pi^2$ gives the sum of the reciprocals of the positive square integers.

$$\sum_{n=1}^\infty\frac{1}{n^2} = \pi^2/6$$

No, there is no general formula yet to know the sum of

$$S(p)=\sum_{n=1}^{\infty} \frac{1}{n^p}$$

But we know that it converges iff $p>1$

For every $p$ even we know that:

$$S(p)=(-1)^{\frac p2+1}{{B_p(2\pi)^p}\over {2p!}}$$

where $B_n$ is a Bernoulli number.

However for odd integers we still don't have a general formula, if you are interested in this kind of series search about Riemann's Zeta function which is a very big topic in math.

More in general if you want to find the value of a random serie you have different convergence tests you can try (like Dirichlet's or Cauchy's one) and then you have to be smart and rearrange the terms in a way you can get out a result, like using telescopic series or Taylor expansions of some well-known functions.

It is, in general, very hard to prove there is NO closed form expression for the sum of a series when the terms of the series are given by a formula. It is much easier, for example, to prove that a closed form formula function has no closed form formula anti-derivative.

On the other hand, there is no reason why we shouldn't be able to (eventually) come up with a closed-form expression for many series sum for many series that we currently can't. This is because both the collection of series whose terms can be expressed in closed-form formula has the same "size" (namely countable) as the number of series whose sum can be expressed in closed form expression. On the other hand, there are uncountably many (i.e. more than countable) convergent series even if we restrict terms to be rational.

When binomial coefficients and such are involved in terms of a series, and also in some other cases, hyper-geometric expressions can often offer a closed-form solution (if you accept that a hyper-geometric expression is a closed form). Manipulations of such series to arrive at closed-form expression for the answer are often tedious, so tools like wolphram alpha are often convenient.

• Without a clear definition of "closed-form", there is nothing to prove. What can be done for antiderivatives is to prove there is no elementary antiderivative. – Robert Israel Sep 3 '15 at 15:19

This is a famous sum & there are multiple ways of approaching this. Here is one way via a geometric sum:

We have the well known sum

\begin{equation*} 1+x+x^2+\cdots =\frac{1}{1-x}. \end{equation*}

Setting $x=e^{i\theta},~0<\theta<2\pi~(x\neq 1)$ gives

\begin{equation*} 1+e^{i\theta}+e^{2i\theta}+\cdots =(1-e^{i\theta})^{-1}=\frac{1}{2}+\frac{1}{2}i\cot(\frac{1}{2}\theta). \end{equation*}

Equating the real parts we see that

\begin{equation*} \frac{1}{2}+\cos(\theta)+\cos(2\theta)+\cdots =0. \end{equation*}

Replacing $\theta$ with $\theta+\pi$ & using the trigonometric addition formulae we get

\begin{equation*} \frac{1}{2}-\cos(\theta)+\cos(2\theta)-\cdots =0. \end{equation*}

Integrate from $\theta=0$ to $\theta=\phi$ & write $\theta$ again for $\phi$ we obtain

\begin{equation*} \sin(\theta)-\frac{1}{2}\sin(2\theta)+\frac{1}{3}\sin(3\theta)-\cdots =\frac{1}{2}\theta,~-\pi<\theta<\pi \end{equation*}

& this series is convergent. Integrating again gives

\begin{equation*} 1-\cos(\theta)-\frac{1-\cos(2\theta)}{2^2}+\frac{1-\cos(3\theta)}{3^2}-\cdots =\frac{1}{4}\theta^2. \end{equation*}

Setting $\theta=\pi$ gives

\begin{equation*} 1+\frac{1}{3^2}+\frac{1}{5^2}+\cdots =\frac{1}{8}\pi^2. \end{equation*}

Since

\begin{equation*} 1+\frac{1}{3^2}+\frac{1}{5^2}+\cdots =1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots -\frac{1}{2^2}-\frac{1}{4^2}-\cdots =(1-\frac{1}{4})(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots), \end{equation*}

we deduce that

\begin{equation*} 1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots = \sum^{\infty}_{n=1} \frac{1}{n^2} =\frac{1}{6}\pi^2. \end{equation*}

Remark: We also get the formula

\begin{equation*} 1-\frac{1}{2^2}+\frac{1}{3^2}-\cdots =\frac{1}{12}\pi^2 \end{equation*}