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I started learning series in calculus and I have trouble catching a basic concept. When I try to find if a series converges or diverges I have many ways to go about it. If I see that the series diverges than I stop there. If I see that the series converges than there is a number it's converging to right?

For example: $\sum\frac 2 {n^3+4}$. I do the limit comparison test with the series $\sum\frac 1 {n^3}$ and get a finite number $2$. I know that that $\sum\frac1{n^3}$ converges, so now I know that $\sum\frac2{n^3+4}$ converges also. How do I know to what number it converges to?

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  • $\begingroup$ I don't know about that particular series, but sometimes this can be really hard. See, for example, some of Ramanujan's series. $\endgroup$ – dfeuer Jan 14 '14 at 14:45
  • $\begingroup$ don't leave out summation signs (unless you're dealing with a context where Feynman's doubled index convention applies). $\sum \cdots$ is a series; $(\cdots)$ is a sequence. $\endgroup$ – dfeuer Jan 14 '14 at 14:50
  • $\begingroup$ Maybe you can use some tricks shown here: youtube.com/watch?v=E-d9mgo8FGk where the sum of all natural numbers converges to -1/12 (it's a headache) $\endgroup$ – CAGT Jan 14 '14 at 14:58
  • $\begingroup$ there is no easy way, for the examples you gave, complex integration is a possible way. $\endgroup$ – Lost1 Jan 14 '14 at 15:05
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    $\begingroup$ nice question with nice answer. $\endgroup$ – Dutta Jan 14 '14 at 15:28
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Other than computing your series numerically (which is usually unsatisfying when it comes to representing a series in a "nice" way with convenient/aesthetic constants like $e,\pi,i,\gamma,$ etc.), the results of most convergence/divergence tests is usually just that: a truth value about whether the thing converges or diverges. For example, the Ratio Test almost always tells you about the nature of convergence, but nothing about it is designed to reveal what the series converges to.

Unfortunately, it is not trivial at all to determine what a series converges to. Before I introduce an example, it is worth our time to make sure everyone is on the same page. Out of all the types of series that you will encounter in your study of series, a "$p$-series" are probably the easiest ones to determine the nature of convergence/divergence. A $p$-series is in the form $$ \zeta(p)= \sum\limits_{k=1}^{\infty} \frac{1}{k^p} $$ Some of you should recognize that the above series is really just the Reimann zeta function. It is worth mentioning what happens to $\zeta(p)$ when $p$ takes on certain values. If $p = 1$, we have what is known as the Harmonic Series which has very wide-ranging applications, particularly in music/acoustics. It turns out that even though the value decrease for each term in $\zeta(1) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots$, the series does not converge. Several proofs can be found here, here, or here. The last source contains a nice poem by Jakob Bernoulli regarding the divergence of $\zeta(1)$.

Anyway, it can also be shown that the only values of $p$ for which $\zeta(p)$ converges are when $p>1$. Now time for an example. Consider a problem (known as the Basel Problem) that stumped many people for many years (until Euler came along [as usual]): We can show readily that $p$-series converge when $p>1$. So a simple case is when $p=2$. What, then, does $$\zeta(2) = \sum\limits_{k=1}^{\infty} \frac{1}{k^2} = ?$$ converge to? I don't want to spoil anything here for you or other curious readers. If you read this post and know the answer, please don't comment/allude to what the answer is right away. Let each reader(s) play around with it for a while and maintain the motivation to do some research on their own. After you can understand that $\zeta(2)$ converges to a somewhat famous closed-form expression, consider how you would determine a similar series for $\zeta(3), \zeta(4), \zeta(5),$ or even $\zeta(n)$ for some arbitrary $n>1$. You'll soon realize that even such innocent-looking, we're-certain-it-will-converge series do not lend themselves easily to obvious answers, let alone closed-form ones!

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  • $\begingroup$ I appreciate your answer as it made it clear for me that serieses are serious business. I tried to follow along your example but didn't quite understand what your'e asking $\endgroup$ – Reboot_87 Jan 14 '14 at 15:47
  • $\begingroup$ @Reboot_87 I expanded my answer for you with a bit more of an introduction to $p$-series and some (hopefully useful) resources using hyperlinks. Enjoy! $\endgroup$ – Xoque55 Jan 15 '14 at 4:22

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