# Convergence of Series

At university, we are currently introduced in various methods of proving that a series converges. For example the ComparisonTest, the RatioTest or the RootTest. However we aren't told of how to calculate the point of convergence. The only series of which we know the point of convergence is the geometric series.

I believe this to be a non trivial task in general, but there have to be at least some methods to calculate the point of convergence for simple series, haven't there?

If you could point me in the direction of some of these methods I would very much appreciate it.

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 The harmonic series is divergent. – Américo Tavares Nov 25 '10 at 23:20 @Américo: He/She probably meant geometric series. – Nuno Nov 25 '10 at 23:26 uups, thanks, I ment the geometric series and have corrected my post accordingly @Nuno, you are absolutly right, thanks ;-) – ftiaronsem Nov 25 '10 at 23:27 What is the "point of convergence"? – Qiaochu Yuan Nov 25 '10 at 23:32 @Qiaochu: (S)He means the limit of the series. – Arturo Magidin Nov 25 '10 at 23:38
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The short answer is that there are no general methods and we use whatever's available. The first thing you can do is learn the Taylor series of common functions; then evaluation of the Taylor series at a point where the series converges to the function corresponds to evaluation of the function. For example, $e^x = \sum \frac{x^n}{n!}$, hence $e = \sum \frac{1}{n!}$. Less trivially, $\ln (1 + x) = \sum \frac{(-1)^{n-1} x^n}{n}$, hence

$$\ln(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} \pm ....$$

The more Taylor series you know, the more series you can evaluate "explicitly" in some sense.

Beyond that, life gets difficult. Mastering the use of Taylor series is already highly nontrivial - especially recognizing when the method is applicable - so worry about learning how to do that properly first.

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Thanks you very much for your answer. Spending some time on TaylorSeries, might indeed prove usefull, not just for analysis but also for physics (which I am studying). – ftiaronsem Nov 26 '10 at 17:51
@ftiaronsem: Not "might"; series expansions are used a lot, implicitly or explicitly, in physics formulae. – Ｊ. M. Nov 27 '10 at 0:45

"there have to be at least some methods to calculate the point of convergence for simple series, haven't there?"

If you are sure a given sequence of numbers converges, you have to just keep iterating everything manually, unless you can think of a clever trick. I suggest you study how Euler proved that $\zeta(2)=\frac{\pi^2}{6}$, that will be very instructive for you because it will demonstrate how clever you have to sometimes be in order to solve this type of problem.

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 Wikipedia tells Euler's approach to the Basel problem. – Luke May 13 '11 at 18:33

Right after the geometric series, the easiest method I know of to calculate the sum of a series is the property of derivation term-by-term of convergent power series. I don't know if you've already covered it in the lessons, so I'll try to give you an idea of how it works:

You know that $\sum x^n=\frac{1}{1-x}$ if $x<1$; now, you will see that you can derive a series like the one before with respect to $x$, and you can do it term-by-term so that you obtain (by differentiating both sides of the previous identity): $\frac{d}{dx}\sum x^n=\sum \frac{d}{dx}x^n=\sum nx^{n-1}=\frac{1}{(1-x)^2}$.

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 Thanks, this also looks very handy. – ftiaronsem Nov 26 '10 at 18:08

There are some convergent series whose exact sum can be evaluated. For instance:

$$\displaystyle\sum_{n=1}^{\infty}\dfrac{n}{(n+1)!}.$$

Observing that

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

the series telescopes. Therefore,

$$\displaystyle\sum_{n=1}^{\infty}\dfrac{n}{(n+1)!}=1-\underset{n\rightarrow\infty}{\lim}\dfrac{1}{(n+1)!}=1-0=1.$$

But in general it is a difficult task to find the exact sum, as you said.

Added: Unlike $\displaystyle\sum_{n=1}^{\infty }\frac{1}{n^{2}}$ whose sum equals $\dfrac{\pi ^{2}}{6}$, the series $\displaystyle\sum_{n=1}^{\infty }\dfrac{1}{n^{3}}$, although convergent, nobody knows a closed form in terms of other mathematical constants. This sum is therefore a mathematical constant in itself.

Added 2: Another example that uses integration and differentiation of the geometric series $$\sum_{n=1}^{\infty }x^{n}=\frac{x}{1-x},\qquad\left\vert x\right\vert <1$$ is

$$\sum_{n=1}^{\infty }\frac{n}{(n+1)e^{n}}=e\left( \log \frac{e-1}{e}+\frac{1% }{e-1}\right) .$$

$$\sum_{n=1}^{\infty }x^{n}=\frac{x}{1-x}\qquad\left\vert x\right\vert <1,$$

we get for $\left\vert x\right\vert <1$

$$\int \sum_{n=1}^{\infty }x^{n}dx=\sum_{n=1}^{\infty }\frac{1}{n+1}% x^{n+1}=\int \frac{x}{1-x}dx=-x-\log \left\vert 1-x\right\vert .$$

Hence,

$$\sum_{n=1}^{\infty }\frac{1}{n+1}x^{n}=-1-\frac{1}{x}\log \left\vert 1-x\right\vert$$

Now if we diferentiate, we have

$$\sum_{n=1}^{\infty }\frac{n}{n+1}x^{n-1}=\frac{1}{x^{2}}\log \left\vert 1-x\right\vert -\frac{1}{x^{2}-x},$$

or equivalently

$$\sum_{n=1}^{\infty }\frac{n}{n+1}x^{n}=\frac{1}{x}\log \left\vert 1-x\right\vert -\frac{1}{x-1}\qquad\left\vert x\right\vert <1.$$

Finally for $x=e^{-1}$, we obtain

$$\sum_{n=1}^{\infty }\frac{n}{(n+1)e^{n}}=e\left( \log \frac{e-1}{e}+\frac{1% }{e-1}\right) .$$

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 Thanks for your answer, but unfortunatelly I can´t follow you on the last part in which you say $\sum_{n=1}^{\infty }\frac{n}{(n+1)e^{n}}=e\left( \log \frac{e-1}{e}+\frac{1% }{e-1}\right)$. Can you be a little bit more specific? Thanks in advance. – ftiaronsem Nov 26 '10 at 18:07 @ftiaronsem, I've now added the evaluation of this sum. – Américo Tavares Nov 26 '10 at 19:09 Wow, thanks for the detailed explanation. Now you helped me twice, both with excellent answers. Thank you very much :-) – ftiaronsem Nov 26 '10 at 22:11