Convergence/Divergence of some series

Question

If I have $\sum_{n=2}^{\infty} \frac {1}{n\log n}$ and want to prove that it diverges, can I use following?

$$\frac {1}{n\log n} \lt \frac {1}{n}$$

• $\sum_{n=1}^\infty \frac 1 n$ diverges, but the limit of $\frac 1 n$ equals to zero so the comparasion I think isn't useful.

• Or can I say it diverges because $\sum_{n=1}^\infty \frac 1{\log n}$ also diverges?

And If I have $\sum_{n=2}^\infty \frac {(-1)^n}{(n)}$ I can use the Leibniz rule. $\sum_{n=1}^\infty \frac 1 n$ diverges, but the limit of it equals to zero, so I am confused if this series diverges/converges again. I understand that $\sum_{n=1}^\infty \frac 1 n$ has an infinite sum, but what to do in this case?

Answer 1

I thought it might be instructive to present an approach that relies on the elementary inequality

$$\log(1+x)\le x \tag1$$

and straightforward arithmetic. To that end, we now proceed.

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From $(1)$ we have

$$\begin{align} \frac{1}{n\log(n)}&\ge \frac{\log\left(1+\frac1n\right)}{\log(n)}\\\\ &\ge \log\left(1+\frac{\log\left(1+\frac1n\right)}{\log(n)}\right)\\\\ &=\log\left(\frac{\log(n+1)}{\log(n)}\right)\tag 2 \end{align}$$

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We can now exploit a telescoping series by using $(2)$ to write

$$\begin{align} \sum_{n=2}^N \frac{1}{n\log(n)}&\ge \sum_{n=2}^N \log\left(\frac{\log(n+1)}{\log(n)}\right)\\\\ &=\sum_{n=2}^N \left(\log(\log(n+1))-\log(\log(n)) \right)\\\\ &=\log(\log(N+1))-\log(\log(2))\tag 3 \end{align}$$

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Finally, letting $N\to \infty$ in $(3)$ yields the coveted result

$$\lim_{N\to \infty}\sum_{n=2}^N \frac{1}{n\log(n)}=\infty$$

and the series of interest diverges.

Answer 2

Your comparison goes in the wrong direction: You need $\dfrac 1 {n\log n} \ge \text{something.}$

Alternatively, you can use an integral test: $$ \int_2^\infty \frac{dx}{x\log x} = \int_2^\infty \left(\frac 1 {\log x} \right) \left( \frac{dx} x\right) = \int_{\log 2}^\infty \frac 1 u\, du = +\infty. $$

Answer 3

Integral test

$$ \int_{2}^{\infty}\frac{1}{x\log(x)}\mathbb{d}x $$

$$ u=\log(x) $$

$$ \mathbb{d}u=\frac{1}{x}\mathbb{d}x $$

$$ \int_{\log(2)}^{\infty}\frac{1}{u}\mathbb{d}u=\lim_{a\to\infty}\left[\log|a|-\log|\log(2)|\right]=\infty $$

So the sum must diverge as well.

Answer 4

Bertrand's series $\displaystyle \sum\limits_{n\ge 2}\frac 1{n^\alpha\ln(n)^\beta}$ converges only for $(\alpha>1)$ or $(\alpha=1,\beta>1)$.

This is generally proved using the comparison to an integral as other answers have shown.

But in general for series of the kind $\displaystyle \sum\limits_{n\ge 2}\frac 1{n^{\alpha_0}\ln(n)^{\alpha_1}\ln(\ln(n))^{\alpha_2}\ln(\ln(\ln(n)))^{\alpha_3}...}$

Which all share the same property of converging only if $\alpha_i=1$ for $i<k$ and $\alpha_k>1$ (with $k$ being the last one), you can use the condensation Cauchy test.

The criterion is : For $(a_n)_n \searrow\ \ge 0$ then $\displaystyle S=\sum\limits_{n\ge 1} a_n<+\infty\iff T=\sum\limits_{n\ge 0} 2^na_{2^n}<+\infty$

with inequality $S\le T\le 2S$.

Applied to $\displaystyle S=\sum\limits_{n\ge 2}\frac 1{n\ln(n)}$, it gives $\displaystyle T=\sum\limits_{n\ge 1}\frac{2^n}{2^n\ln(2^n)}=\frac 1{\ln(2)}\sum\limits_{n\ge 1} \frac 1n$ which is divergent.

Answer 5

Since $$ \sum_{n=2}^\infty\frac{2^n}{2^n n\log(2)} $$ diverges by comparison to the Harmonic Series, the Cauchy-Condensation Test says that $$ \sum_{n=2}^\infty\frac1{n\log(n)} $$ also diverges.

Answer 6

The really long explanation of what robjohn's answer and Gareth McCaughan's answer are trying to say for those who like it written out.

Perhaps it will help to see what the Cauchy condensation test says:

$$f(n)\ge f(n+k)\implies\\\sum_{n=1}^\infty2^nf(2^{n+1})\ge\sum_{n=2}^\infty f(n)\ge\sum_{n=1}^\infty2^{n-1}f(2^n)$$

Here, we have $f(n)=\frac1{n\log(n)}$. A visualization of what's happening:

$$\small\begin{align}\sum_{n=2}^\infty\frac1{n\log(n)}&=\frac1{2\log(2)}+\frac1{3\log(3)}+\frac1{4\log(4)}+\frac1{5\log(5)}+\frac1{6\log(6)}+\frac1{7\log(7)}+\frac1{8\log(8)}+\dots\\&\ge\underbrace{\frac1{2\log(2)}}_{2^0f(2^1)}+\underbrace{\frac1{4\log(4)}+\frac1{4\log(4)}}_{2^1f(2^2)}+\underbrace{\frac1{8\log(8)}+\frac1{8\log(8)}+\frac1{8\log(8)}+\frac1{8\log(8)}}_{2^2f(2^3)}+\dots\\&=\sum_{n=1}^\infty2^{n-1}f(2^n)\end{align}$$

It is then noted that:

$$2^{n-1}f(2^n)=\frac1{2n\log(2)}$$

And as we know that $\sum_{n=1}^\infty\frac1n$ diverges, we can see that

$$\sum_{n=2}^\infty\frac1{n\log(n)}\ge\sum_{n=1}^\infty2^{n-1}f(2^n)=\frac1{2\log(2)}\sum_{n=1}^\infty\frac1n=+\infty\\\implies\sum_{n=2}^\infty\frac1{n\log(n)}=+\infty$$

Also note that your comparison is done backwards. It is always the case that you can find an upper bound that sums to infinity, but to get results from the comparison test, you need to find a lower bound that sums to inifnity, as we've done here.

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As for the Leibniz rule and the alternating harmonic series, notice that:

$$\small\begin{align}\sum_{n=1}^\infty\frac{(-1)^{n+1}}n&=\frac11-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17-\frac18+\dots\\&\le\frac11-\frac12+\frac12-\frac14+\frac14-\frac16+\frac16-\frac18+\dots\\&=1\end{align}$$

$$\small\begin{align}\sum_{n=1}^\infty\frac{(-1)^{n+1}}n&=\frac11-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17-\frac18+\dots\\&\ge\frac11-\frac11+\frac13-\frac13+\frac15-\frac15+\frac17-\frac17+\dots\\&=0\end{align}$$

Thus, we have

$$0\le\sum_{n=1}^\infty\frac{(-1)^{n+1}}n\le1$$

And as $\lim_{n\to\infty}\frac{(-1)^{n+1}}n=0$, it must be the case that the alternating harmonic series converges, as it is bounded and the difference between partial sums tends to zero.

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On the other hand, the harmonic series is not bounded above. There are plenty of approaches to showing this. Using the Cauchy condensation test from above, we find that

$$f(n)=\frac1n\\\begin{align}\sum_{n=1}^\infty\frac1n&=\sum_{n=1}^\infty f(n)\\&\ge1+\sum_{n=1}^\infty2^{n-1}f(2^n)\\&=1+\sum_{n=1}^\infty\frac{2^{n-1}}{2^n}\\&=1+\sum_{n=1}^\infty\frac12\\&=1+\frac12+\frac12+\frac12+\dots\\&=+\infty\end{align}$$

Answer 7

Just for fun, here's another proof, inspired by the usual elementary proof that $\sum\frac{1}{n}$ diverges. As with that proof, split into sums over integers {2,3}, {4,5,6,7} and so on: $2^k\leq n<2^{k+1}$. Within each such range, $\frac1{n\log n}$ changes by a factor no bigger than 4, because each of $n$, $\log n$ changes by a factor no bigger than 2. So our sum is within a factor of 4 of $\sum_k 2^k\frac{1}{2^k\log2^k}$ which, again changing by only a constant factor by taking the logs to be to base 2, equals $\sum_k\frac1k$ ... and we know how that behaves.

We can iterate this construction to show that the sum of $\frac1{n\log n\log\log n}$ diverges, and so do the similar series with more and more factors in the denominator.

[EDITED to add:] This is of course more or less the same answer as robjohn's, but with its logic written out explicitly; and robjohn's is more or less the same as zwim's. And in some sense these are all "morally" the same as the integral-based argument in the accepted answer, with its change of variable $u=\log x$; what zwim, robjohn and I are all doing is a similar logarithmic/exponential change of variable while staying in the realm of the discrete.