I'm finding it quite difficult to understand the idea of series and limits to test for divergence or convergence. Perhaps more so in finding such a limit.
I have the series
$$\sum_{n=1}^\infty \frac{1}{2n}$$
I just have no idea... so any points in the right direction would be appreciated.
There's a theorem in my book for "Test for Divergence" stating that if $\lim_{n\rightarrow\infty}a_n$ does not exist, or does not equal zero then the series is Divergent.
I calculated $\lim_{n\rightarrow\infty}\frac{1}{2n} = \lim_{n\rightarrow\infty}\frac{1/n}{2} = 0$
So what does this mean? Apparently if a series is convergent the limit is always zero. But it does not necessarily mean that if the limit of a series is zero that it is convergent.
So how can I work it out?
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$\begingroup$Whether or not a series converges does not depend on if $\lim_{n\to\infty}a_n=0$, rather it is the converse that is true.
If a series converges, then $\lim_{n\to\infty}a_n=0$.
But $\lim_{n\to\infty}a_n=0$ does not prove that a series converges.
Your series is an example of one that does not converge and approaches $0$.
Here is my explanation:
If a sum gets closer and closer to some limit, then its partial sums must increase by less and less. Upon reaching the limit, the partial sums no longer increase and we must have the following:$$\lim_{n\to\infty}\sum_{i=1}^{n}{f(i)}=\lim_{n\to\infty}\sum_{i=1}^{n+1}{f(i)}$$Due to the nature of $\infty$.
Furthermore, we have such:
$$\lim_{n\to\infty}\sum_{i=1}^{n}{f(i)}=\lim_{n\to\infty}\sum_{i=1}^{n+1}{f(i)}$$$$\lim_{n\to\infty}f(1)+f(2)+f(3)+\cdots f(n)=\lim_{n\to\infty}f(1)+f(2)+f(3)+\cdots f(n)+f(n+1)$$$$0=\lim_{n\to\infty}f(n+1)$$
One can prove a convergent series has this property, but not that a series with this property is convergent.
This means that a series without this property must be divergent, because I just proved all convergent series have this property, meaning if it doesn't, it can't be convergent.
Also, (fun fact) a series' sum is not dependent on whether it converges.
For example:$$\sum_{n=1}^{\infty}n=-\frac12$$If you don't believe, you can look it up. It even has its own wiki page.
$\endgroup$ 2 $\begingroup$As others have said this is not a sufficient condition. For this particular example you can argue as follows. Suppose to the contrary that the series converges.
Let $s_n$ denote the $n$-th partial sum. Since the serie converges then $(s_n)$ is a Cauchy sequence. Let $\varepsilon = 1/6$. So there is $n_0$ such that $|s_q-s_p|< 1/6$ for all $q>p\ge n_0$. Let $q=2n_0$ and $p=n_0$. Then
$$\frac{1}{6}>\bigg|\sum_{n=n_0+1}^{2n_0} \frac{1}{2n}\bigg|\ge\bigg|\sum_{n=n_0+1}^{2n_0} \frac{1}{4n_0}\bigg|=\frac{1}{4}$$
a contradiction. Then this contradiction shows that the series diverges.
$\endgroup$ $\begingroup$$$1/2+1/4+1/6+1/8+1/10+1/12 +1/14 +1/16+......\geq$$ $$\geq 1/2+1/4+(1/8+1/8)+(1/16+1/16+1/16+1/16)+...=$$ $$=1/2+1/4+1/4+1/4+...$$ which is diverging.
There are many tests for convergence. For a monotone sequence like this one the Cauchy Condensation Test is often useful: If $f:\mathbb N\to \mathbb R$ is monotonic then $\sum_nf(n)$ converges iff $\sum_n2^nf(2^n)$ converges.
Here, $f(n)=1/2n$ is monotonic, and $f(2^n)=1/(2\cdot 2^n)=1/2^{n+1}$. So $2^nf(2^n)=2^n/2^{n+1}=1/2 $ and $\sum_n2^nf(2^n)=\sum_n(1/2)=1/2+1/2+1/2+...,$ diverging.
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