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So, I'm a bit baffle on this concept for the convergence of series.

Test for Divergence: http://imgur.com/XZk547T (2/5)≠0 So, it's divergent.

Now, let's take a look at this example: http://imgur.com/EWYc8Dk

For this example, it converges to -(8/7). What? Did I miss something? I thought it had to be zero in order to converge?

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    $s_n$ denotes the sequence of partial sums, i.e. $\displaystyle s_n:=\sum_{k=0}^{n}a_k$. The divergence test deals with $\displaystyle\lim_{k\to\infty}a_k$ – 2017-01-28

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Every series has two sequences associated with it.

If our series is $S = \displaystyle\sum_{n=1}^{+\infty} a_n$, then the two sequences are:

  • $a_1, a_2, a_3, \dots$. This is the sequence of the terms of the series.
  • $S_1, S_2, S_3, \dots$. This is the sequence of partial sums of the series. We define these as $S_k = \displaystyle\sum_{n=1}^k a_n$. For example, $S_3 = a_1 + a_2 + a_3$.

I thought it had to be zero in order to converge

The limit of the sequence must be zero. In other words, $\displaystyle\lim_{n\to+\infty} a_n = 0$ must be true.

The series itself could potentially take on any value. In other words, $\displaystyle \lim_{n\to+\infty} S_n$ is not necessarily zero since a series could converge to a nonzero number.

For example, $\displaystyle \sum_{n=1}^{+\infty} \sin n$ diverges, because $\displaystyle \lim_{n\to+\infty} \sin n$ does not exist (and therefore is not zero).

Another example is $\displaystyle \sum_{n=1}^{+\infty} \frac1{n^2} = \frac{\pi^2}6$.

WARNING: The limit of the sequence being zero is not a sufficient condition for convergence. For example, $\displaystyle \sum_{n=1}^{+\infty}\frac1n$ diverges even though $\displaystyle \lim_{n\to+\infty}\frac1n = 0$.

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    It's starting to make much more sense, but there's something stopping me from completely comprehending it. The weird thing is that I don't have an issue with other problems that involve n!, n^n, b^n (b>1), etc. It's only problems that have polynomials on top and the bottom with equal powers. I'm just too fixated with the idea of limits being able to approach a number and that number being the number it convergences to (every single time). – 2017-01-28
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    @Mathematicu5 if that's the case then I think you'll get the hang of it with a bit more practice. Just keep in mind that there are two sequences associated with every series. Also if the sequence $a_n$ truly is a rational function of $n$ (polynomial on top and polynomial on bottom, nothing else "non-polynomial" like $\sqrt n$ or $\cos n$, etc.) and if the degrees of the numerator and denominator are the same, then the series $\sum_n a_n$ will always diverge because $\lim_{n\to+\infty} a_n$ can never be zero in this case. – 2017-01-28
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    Thank You! Honestly, you just beat the crap of that voice in my head that wasn't allowing me to comprehend this. Thank You! – 2017-01-28
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By the definition of convergence of a series:

$\sum_{n=0}^\infty a_n$ converges if $L=\lim_{N\to\infty}s_N=\lim_{N\to\infty}\sum_{n=0}^N a_n$ exists.

It then follows that if $\lim_{N\to\infty}s_N$ exists, then

$\lim_{N\to\infty}a_N=\lim_{N\to\infty}s_N-s_{N-1}=L-L=0$

Thus, for a series to converge, each individual term must approach zero.

However, the converse is not true. Just because each term goes to zero does not mean the series will converge.

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    Just $\sum_{n=0}^{\infty} a_n$, not $\lim_{n\to\infty} \sum_{n=0}^\infty a_n$. – 2017-01-28
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    @tilper Oops, thanks for the catch, I fixed it. – 2017-01-28