# How to prove $\sum\limits_{k=1}^{\infty} {\sqrt\frac{1}{\pi k}} = \sqrt\frac{1}{\pi}\sum\limits_{k=1}^{\infty} {\sqrt\frac{1}{k}} = \infty$ [closed]

My prof mentioned during a proof the following statement is from Calculus but I can't seem to prove it. Any suggestions?

$$\sum\limits_{k=1}^{\infty} {\sqrt\frac{1}{\pi k}} = \sqrt\frac{1}{\pi}\sum\limits_{k=1}^{\infty} {\sqrt\frac{1}{k}} = \infty.$$

I mainly don't understand how $$\sum_{k=1}^{\infty} \frac{1}{\sqrt k} = \infty$$, because $$\lim_{k\to \infty} \frac{1}{\sqrt k} = 0$$, so intuitively speaking the sum of the fraction should approach a set number, not $$\infty$$ ?

• What series do you know diverge? Usually $\sum \frac{1}{\sqrt{k}}$ is not the first divergent series you get introduced to. Feb 28 at 23:06
• If that limit would not approach $0$, then you would know that the series would diverge(this is what "$=\infty$" really means). But the fact that the limit is $0$ doesn't say anything about convergence. Your series is called a p-series. You can learn more about them online, and prove their convergence/divergence with the integral test. Feb 28 at 23:08
• this is actually part of the proof for a simple random walk problem for stochastic processes, my prof just used this claim but I can't really understand why. Feb 28 at 23:11
• @Lunedi It is a bit unusual that you are dealing with stochastic processes but have never had to deal with simple divergent series. They should be part of an introductory analysis course, way before one encounters stochastic processes. Do you know that the famous harmonic series $\sum_{k=1}^\infty \frac{1}{k}$ diverges?
– Gary
Feb 28 at 23:42
• Check out this question and its top answer to see why the terms going to $0$ does not necessarily mean that the series is finite. Mar 1 at 1:15

Using the comparison test, we can show that $$\sum_{k = 1} ^\infty \frac{1}{\sqrt{k}}$$ diverges as $$\sum_{k = 1} ^\infty \frac{1}{k}$$ diverges. Note that $$\frac{1}{k} \leq \frac{1}{\sqrt{k}}$$ for all $$k \geq 1$$.

Are you aware of the Condensation test? If not, you can check it out here. Based on it, since $$a_{k} > 0$$ and nonincreasing, the proposed series converges (diverges) iff the following series converges (diverges): \begin{align*} \sum_{k=1}^{\infty}2^{k}a_{2^{k}} = \sum_{k=1}^{\infty}2^{k}\frac{1}{\sqrt{2^{k}}} = \sum_{k=1}^{\infty}2^{k/2} = \sum_{k=1}^{\infty}(\sqrt{2})^{k} \end{align*}

Once the last series diverges, the original series also diverges.

That is because a geometric series diverges when its ratio is greater than one.

Hopefully this helps !

The first step is essentially that $${\sqrt\frac{1}{\pi k}}={\sqrt\frac{1}{\pi }}{\sqrt\frac{1}{ k}}$$ since $$\frac 1 \pi$$ and $$\frac1k$$ are positive, and then taking the constant $$\sqrt\frac{1}{\pi }$$ outside the summation. It is not particularly interesting here.

There are several different ways to show $$\sum\limits_{k=1}^{\infty} {\sqrt\frac{1}{k}}$$ is infinite. One is to say $$\sqrt\frac{1}{x}$$ is a decreasing function of $$x$$ with integral $$2\sqrt{x}+c$$ so $$\sqrt\frac{1}{k} \ge \int\limits_{x=k}^{k+1} \sqrt\frac{1}{x} \,dx$$ and thus $$\sum\limits_{k=1}^{n} {\sqrt\frac{1}{k}} \ge \int\limits_{x=1}^{n+1} \sqrt\frac{1}{x} \,dx =2\sqrt{n+1}-2$$ and, since the right hand side increases towards infinity as $$n\to \infty$$, the left hand side which is larger must too. You can use a similar argument to say the partial sum is no more than $$2\sqrt{n}-1$$, so you now also know how fast it increases towards infinity.

Since $$1/\sqrt{k}$$ is a decreasing function, we can say that $$\sum_{k=1}^{N}\frac{1}{\sqrt{k}} \ge \int_{1}^{N+1}\frac{dk}{\sqrt{k}}=2\sqrt{k}\Big\vert_{1}^{N+1}=2\sqrt{N+1}-2.$$ This diverges to $$\infty$$ as $$N\rightarrow\infty$$, and so the original series also diverges.