Showing that the Series $\frac{1}{n^k}$ converges

Question

I am having problems this time in proving that the following series converges for all $k > 1$. k is also element of the natural numbers (k e N, someone please edit this correctly)

Note: I should not use the integral test with this, so please don't use it.

$$\sum_{n=1}^\infty = \frac{1}{n^k}$$

My approach to this using the ratio test:

$$ \frac{\frac {1}{(n+1)^k}}{\frac{1}{n^k}} = \frac{n^k}{(n+1)^k} $$

And now I can say:

$$\frac{n^k}{(n+1)^k}<1 $$

Because $(n+1)^k$ is clearly greater than $n^k$. So this Series converges for all k>1.

But I have the feeling that this is clearly not enough as a proper answer to this task, since it gives 2/20 points and this simply seems to easy and just not "right".

Best regards,

SacredScout

Answer 1

I thought it might be instructive to present an approach that forgoes appealing to tests, such as the integral test and Cauchy's condensation test. Instead, we use elementary analysis including telescoping series and Bernoulli's Inequality. To that end we now proceed.

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We begin with the telescoping series

$$\sum_{n=1}^N \left(\frac1{n^s}-\frac1{(n+1)^s}\right)=1-\frac1{(N+1)^s}\tag1$$

In addition we have from Bernoulli's Inequality for $0< s$

$$\begin{align} \frac1{n^s}-\frac1{(n+1)^s}&=\frac1{n^s}\left(1-\frac1{(1+1/n)^s}\right)\\\\ &=\frac1{n^s}\left(1-\frac{1+\frac1n}{(1+1/n)^{s+1}}\right)\\\\ &\ge \frac1{n^s}\left(1-\frac{1+\frac1n}{1+\frac{s+1}{n}}\right)\\\\ &\ge \frac{s}{(s+2)n^{s+1}}\tag2 \end{align}$$

Using $(1)$ and $(2)$ reveals that for $s>0$

$$\sum_{n=1}^N \frac{1}{n^{s+1}}\le \frac{s+2}{s}\left(1-\frac1{(N+1)^s}\right)\tag3$$

Inasmuch as the sequence of partial sums $\sum_{n=1}^N \frac{1}{n^{s+1}}$ is increasing and bounded by the term of the right-hand side of $(3)$, it converges.

And we are done!

Answer 2

We can group terms and bound it by a convergent geometric series.

\begin{align} \sum_{n=1}^\infty \frac1{n^k} &= \sum_{m=0}^\infty \sum_{j=2^m}^{2^{m+1}-1} \frac1{j^k} \\ &\le \sum_{m=0}^\infty \frac{2^{m+1}-2^m}{2^{mk}}\\ &=\sum_{m=0}^\infty \frac{2^m}{2^{mk}}\\ &=\sum_{m=0}^\infty\left( \frac{2}{2^k}\right)^m\\ &= \sum_{m=0}^\infty \left(\frac1{2^{k-1}} \right)^m\\ &= \frac{1}{1-\frac1{2^{k-1}}} \end{align}

Note that for ratio test, we have to take the limit and in this case, the limit is $1$.

Answer 3

HINT

Unfortunaltely ratio test is not conclusive since $\frac{\frac {1}{(n+1)^k}}{\frac{1}{n^k}} = \frac{n^k}{(n+1)^k}\to 1$.

The standard way is to use integral test

$$\int_{1}^\infty \frac1{x^k}dx\le \sum_{n=1}^\infty\frac{1}{n^k} \le 1+\int_{1}^\infty \frac1{x^k}dx$$

proving the convergence case by RHS inequality and the divergence case by LHS inequality.

As an alternative by Cauchy condensation test comapring the series with the condensed series for $k>0$

$$\sum_{n=1}^\infty\frac{2^k}{(2^k)^k}=\sum_{n=1}^\infty\frac{1}{2^{k(k-1)}}$$

Answer 4

Firstly, note that your proof must be wrong, because if it weren't, we could simply take $k = 1$, go through the same process, and prove that the harmonic series converges, which is not true. The error, specifically, is at the end: the ratio test requires more than that the ratio is less than $1$ for all $n$: it requires that there is some $l < 1$ (constant for all $n$) such that the ratio is less than $l$ for all $n$. None of these sequences satisfy this, since $\left(\dfrac{n^k}{(n+1)^k}\right)\to 1$ as $n \to \infty$.

Instead, you can use any of the methods here (which it's very unlikely that anybody expected you to produce on your own, but the second (by Harald Hanche-Olsen) is probably the one that's most reasonable), or the Cauchy Condensation Test.