Proving the series $s_n=1+\frac{1}{\sqrt2}+\cdots +\frac{1}{\sqrt n}$ is not bounded

Question

Let $s_n=1+\frac{1}{\sqrt2}+\cdots +\frac{1}{\sqrt n}$. This sequence is definitely increasing, and I tried to show that it is not bounded above, by induction. However if $s_n\leq2$, then $s_{n+1}\leq2+\frac{1}{\sqrt n}\nleq2$. So I thought this is not bounded above.

Now, the second sequence is $s_n=\frac{1}{1^2}+\frac{1}{2^2}+\cdots+\frac{1}{n^2}$. The problem requires to prove that this one is bounded above by $2$.

I tried to apply the same method as before: if $s_n\leq2$, then $s_{n+1}\leq2+\frac{1}{(n+1)^2}\nleq2$. But, this is bounded above by $2$.

What is wrong with my logic?

Answer 1

Your argument indeed shows that $s_{n+1}\le 2+ \dfrac{1}{\sqrt{n+1}}$, but that doesn't mean that $s_{n+1} > 2$ since if you have $s_{n+1}\le2$ then you certainly have $s_{n+1}\le 2+ \dfrac{1}{\sqrt{n+1}}$.

As a hint for your problem, try starting with $s_n^2$ along with the fact that any sequence $a_n$ is bounded if and only if $\sqrt{a_n}$ is bounded.

Answer 2

What is wrong with your logic?

Well, if $s_n\le 2$ and $s_{n+1}=s_n+c$ with $c\ge0$, that means $s_n\le s_{n+1}$,

but it does not mean $2\le s_{n+1}$.

In fact, for the second sequence, $s_1=1\le2$, and $s_2=s_1+\frac14=\frac54<2.$

Answer 3

The statement

$$s_n\le2\implies s_{n+1}\le 2+\frac1{\sqrt n}\not\le 2$$

is

• not justified,

• false.

(Take $s_1=1$.)

But, what's worse, $>2$ does not mean unbounded at all

Answer 4

Hint: consider the integral $S_n = \int_1^n \frac{dx}{\sqrt{x}}$. Note that $(\sqrt{x})' = \frac 1 2 \frac{1}{\sqrt x}$ so your integral is infinite. Because the integrand is decreasing, you can compare $S_n$ and $T_n = \sum_{k=1}^n \frac{1}{\sqrt{k}}$. This will actually show that for any positive strictly decreasing $f$, the integral

$$\int_1^\infty f(t)dt$$

is finite if and only if the infinite sum

$$\sum_{k=1}^\infty f(k)$$ is finite.

Answer 5

You are mixing some ideas here. If you assume $s_n\le2$ and deduce $s_{n+1}\le2+\frac{1}{\sqrt{n+1}}$, this logic is correct (if you prove it for the induction base $s_1$ - this is called proof by induction). However, you did not prove that $s_{n+1}\ge 2$ - intuitively, you don't know "how much" smaller from $2$ is $s_n$, so adding $\frac{1}{\sqrt{n+1}}$ to it might still make it smaller than $2$, so you did not prove that it is not bounded. As for your question, try looking at the answer given here.

For your second question, try using this inequality (which you should prove rigorously using induction): $$ \begin{align} &\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2} +\frac{1}{5^2} +\frac{1}{6^2} +\frac{1}{7^2} +\frac{1}{8^2} +⋯ \\ &<\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{2^2}+\frac{1}{4^2} +\frac{1}{4^2} +\frac{1}{4^2} +\frac{1}{4^2} +\frac{1}{8^2}+\dots \\ &=1+2⋅\frac{1}{2^2} +4⋅\frac{1}{4^2}+8⋅\frac{1}{8^2} +⋯ \\ &=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+⋯ \\ &=2. \end{align}$$

[Side note - this is a very interesting sequence. The infinite sum was unknown for more than a century, until Euler proved it is equal to the surprising value of $\frac{\pi^2}{6}$ in 1734. This problem was called Basel problem. Also check the very nice 3Blue1Brown video.]