Analysis: Proving divergence using partial sums

Question

How do I prove that $$\sum_{n=1}^{\infty}\frac{1}{n^s}$$ diverges for $s<1$, by estimating its partial sums?

Answer 1

For $s=1$, this series is called the harmonic series, and we can prove directly that it diverges. Note that $\frac{1}{3}+\frac{1}{4}>\frac{1}{2}$, that $\frac{1}{5}+\space ...\space+\frac{1}{8}>4\frac{1}{8}=\frac{1}{2}$, and in general that $$\frac{1}{2^{k-1}+1}+\frac{1}{2^{k-1}+2}+\space ...\space+\frac{1}{2^k}\space >2^{k-1}\frac{1}{2^k}=\frac{1}{2}.$$

Hence, the $(2^k)^{th}$ partial sum $S_{2^k}$ satisfies $S_{2^k}>1+k\frac{1}{2}$. Since the terms in the harmonic series are all positive, the sequence of partial sums is monotonically increasing, and by the calculation done the sequence of partial sums is unbounded, and so the sequence of partial sums diverges. Hence, the harmonic series diverges.

Note that for $s<1$, we have that $n^s<n$ (even for $s=0$ or $s$ negative), and hence that $\frac{1}{n^s}>\frac{1}{n}$. Hence, if we let $S_k$ be the $k^{th}$ partial sum of the harmonic series $\sum_{n=1}^{\infty}\frac{1}{n}$, and $T_k$ be the $k^{th}$ partial sum of the series $\sum_{n=1}^{\infty}\frac{1}{n^s}$ under consideration, then $T_k>S_k$. Since $\frac{1}{n^s}>0$ for all $n$, we have that $T_k$ is an unbounded monotonically increasing sequence, unbounded since $S_k$ is unbounded by the argument given in the paragraph above, and so $T_k$ diverges. So, by definition, $\sum_{n=1}^{\infty}\frac{1}{n^s}$ diverges.

Answer 2

It was a long time since I did this, but I think it would work to do a elementwise estimate $$\frac{1}{n} < \frac{1}{n^s}, s \in ]0,1]$$ and using $$\sum_{n=1}^\infty \frac{1}{n}$$ which diverges (it is the famous harmonic series).

Answer 3

$$\sum_{n=1}^n \frac1{n^s}=1+\sum_{n=2}^n \frac1{n^s}=1+\int_2^{n+1}\frac1{\lfloor x\rfloor^s}dx$$ $$\lfloor x\rfloor<x$$ $$\sum_{n=1}^n \frac1{n^s}>1+\int_2^{n+1}\frac1{x^s}dx=1+\frac{1}{1-s}\left[x^{1-s}\right]_2^{n+1}=1+\frac{(n+1)^{1-s}-2^{1-s}}{1-s}$$ Therefore $\sum_{n=1}^{\infty} \frac1{n^s}$ diverges when $s<1$.

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On the other hand, $$\lfloor x\rfloor\ge x-1$$ $$\sum_{n=1}^n \frac1{n^s}\le1+\int_2^{n+1}\frac1{(x-1)^s}dx=1+\frac{1}{1-s}\left[(x-1)^{1-s}\right]_2^{n+1}=1+\frac{n^{1-s}-1}{1-s}$$ Therefore $\sum_{n=1}^{\infty} \frac1{n^s}$ converges when $s>1$.

Answer 4

Write the sum in a fraction and see what happen with the partial sums:

$$\sum_{n=j}^{N}\frac{1}{n^s}=\frac{1}{j^s}+\frac{1}{(j+1)^s}+...+\frac{1}{N^s}=\frac{\sum_{k=j}^{N}\frac{1}{k^s}\left(\prod_{n=j}^{N} n^s\right)}{\prod_{n=j}^{N} n^s}$$

Now we have that

$$\frac{(N-j+1)\frac{1}{N^s}\left(\prod_{n=j}^{N} n^s\right)}{\prod_{n=j}^{N} n^s}<\frac{\sum_{k=j}^{N}\frac{1}{k^s}\left(\prod_{n=j}^{N} n^s\right)}{\prod_{n=j}^{N} n^s}$$

$$\frac{N-j+1}{N^s}<\sum_{n=j}^{N}\frac{1}{n^s}$$

Then if $\exists N\ \forall j: (j,N\in\Bbb N)\land\left(\frac{N-j+1}{N^s}\ge 1\right)$ then the series is divergent. This is the case for $0<s<1$ because the growth rate of denominator and numerator are different, so you can ever choose some $N$ that make true the above inequality.

Notice that you can prove divergence from this but not convergence i.e. you cant says anything for the cases $s\ge 1$ because the inequality doesnt hold.