Convergent of p-series

Question

My lecturer went through this method for proving $\sum_{n=1}^\infty \dfrac{1}{n^p} $ converges if $p>1$

Since $1/n^p > 0$ it implies that the sequence of partial sums $S_N$ is increasing.

So either $S_n$ is bounded above & converges or it diverges to $+\infty$

Then he went on to say if $\exists$ subseq $(S_{jN})$ that is bounded above: then $S_{jN}$ does not go to $+\infty$ if $n\to \infty$ which implies that $S_N$ does not go to infinity which implies $(S_N)$ converges

So in the proof we showed that a subsequence $S_{2^{N}-1}$ is bounded, and since it's increasing it converges, which is great. HOwever I don't understand the logic here, if a subsequence of $(S_N)$ converges, then why does that mean that $(S_N)$ converges? I'm aware that if $(S_N)$ converges then all subsequences of $(S_N)$ also converge to same limit, but I don't understand this proof as all he showed is that one particular subsequence converges, and then he concludes with $(S_N)$ converges.

If anyone could clear this up.. thank you

score 1 · Accepted Answer · answered Dec 28 '13 at 18:37

1

Let $k$ be the value that bounds $S_{2^N-1}$. Since $S_N$ is increasing, $S_N \leq S_{2^N-1} < k$ for $N>0$. Therefore $S_N$ is bounded.

answered Dec 28 '13 at 18:37

Zach H

903

How did you get $S_N \leq S_{2^N-1} < k$ sorry? – DHx Dec 28 '13 at 18:41
Because $N\le 2^N-1$ and $S_n$ is increasing – Hagen von Eitzen Dec 28 '13 at 19:03

score 1 · Answer 2 · answered Dec 28 '13 at 20:40

To understand intuitively why the sequence of partial sums above must converge if it has a bounded subsequence, suppose that it does not converge. Then since it is increasing, it must diverge to +$\infty$. That means that for each $M$, there exists an $n$ for which $S_n > M$. Since $(S_i)$ is increasing, this means that $S_m$ exceeds $M$ for every $m \ge n$. That makes it impossible for any subsequence of partial sums to stay below $M$.

score 0 · Answer 3 · answered Dec 28 '13 at 18:34

Theorem: Let $(x_n)$ denote a nondecreasing sequence, then $(x_n)$ converges iff $(x_n)$ is bounded above iff every subsequence $(x_{\varphi(n)})$ converges iff there exists a subsequence $(x_{\varphi(n)})$ which converges iff there exists a bounded above subsequence $(x_{\varphi(n)})$ iff...

For example:

If some subsequence $(x_{\varphi(n)})$ is bounded above, say $x_{\varphi(n)}\leqslant c$ for every $n$, then, for every $k$ there exists some $n$ such that $k\leqslant\varphi(n)$, hence $x_k\leqslant x_{\varphi(n)}\leqslant c$, thus $(x_k)$ is bounded above. Now, every nondecreasing sequence bounded above is $______$. QED.

Convergent of p-series

3 Answers3