Find $\lim_{n \rightarrow \infty} \frac{1}{\sqrt{n}}(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....\frac{1}{\sqrt{n}})$ Find $\lim_{n \rightarrow \infty} \frac{1}{\sqrt{n}}(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....\frac{1}{\sqrt{n}})$
I was thinking of Using Cesaro's lemma.
I define: $S_{n}= 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....\frac{1}{\sqrt{n}}$
and $T_{n}=\sqrt{n}$
According to the lemma, if $\lim_{n \rightarrow \infty} \frac{S_{n+1}-S_{n}}{T_{n+1}-T_{n}}$ exists then $\lim_{n \rightarrow \infty} \frac{S_{n}}{T_{n}}$ exists and has the same limiting value. So, I am getting the limit to be $0$ .Am I correct?
 A: Hint (for an alternative approach): think "Riemann sums" (this does not exactly applies as it is right away, IIRC, as $f$ below is not Riemann integrable. But the idea is the same, i.e. to compare your sum to an integral, see at the end.)
$$
\frac{1}{\sqrt{n}}\sum_{k=1}^n \frac{1}{\sqrt{k}}
= \frac{1}{n}\sum_{k=1}^n \sqrt{\frac{n}{k}}
= \frac{1}{n}\sum_{k=1}^n f\left(\frac{k}{n}\right)
$$
where $f\colon x\in (0,1] \mapsto 1/\sqrt{x}$.
(Basically, the underlying idea amounts to comparing the behaviour of the partial sums of the divergent series $\sum_{k=1}^n \frac{1}{\sqrt{k}}$ to the integral $\int_1^n \frac{dx}{\sqrt{x}}$, via a comparison series/integral.)
A: Let $S_n=\frac{1}{\sqrt{n}}(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....\frac{1}{\sqrt{n}})$. Then
$$
\frac{1}{\sqrt{n}}\int_0^{n}\frac{dx}{\sqrt{x}}\geq S_n\geq \frac{1}{\sqrt{n}}\int_0^{n}\frac{dx}{\sqrt{x+1}}.
$$
Therefore
$$
2\geq S_n\geq \frac{2}{\sqrt{n}}(\sqrt{n+1}-1)
$$
and $\lim_{n\to\infty}S_n=2$.
A: Another approach is to use the Abel's summation $$\sum_{k=1}^{n}\frac{1}{\sqrt{k}}=\sum_{k=1}^{n}1\cdot\frac{1}{\sqrt{k}}=\sqrt{n}+\frac{1}{2}\int_{1}^{n}\frac{\left\lfloor t\right\rfloor }{t^{3/2}}dt
 $$ where $\left\lfloor t\right\rfloor 
 $ is the floor function. Since $\left\lfloor t\right\rfloor =t+O\left(1\right)
 $ we have $$\sum_{k=1}^{n}\frac{1}{\sqrt{k}}=\sqrt{n}+\frac{1}{2}\int_{1}^{n}\frac{1}{t^{1/2}}dt+O\left(\int_{1}^{n}\frac{1}{t^{3/2}}dt\right)
 $$ $$=2\sqrt{n}+O\left(1\right)
 $$ so $$\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{\sqrt{k}}=2+O\left(\frac{1}{\sqrt{n}}\right)\rightarrow2
 $$ as $n\rightarrow\infty$.
A: The approach is correct but I think the limit is 2.
A: It is a simple Riemann sum for the function $\dfrac1{\sqrt x}$ on the interval $(0,1]$ It converges to 
$$\int_0^1\frac{\mathrm d\mkern 1mu x}{\sqrt x}=2\sqrt x\,\Bigr\rvert_0^1=2.$$
A: Using $$2\left(\sqrt{k+1}-\sqrt{k}\right)\leq \frac{1}{\sqrt{k}}\leq 2\left(\sqrt{k}-\sqrt{k-1}\right)$$
So $$\lim_{n\rightarrow \infty}2\frac{\left(\sqrt{n+1}-\sqrt{1}\right)}{\sqrt{n}}\leq \lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\sum^{n}_{k=1}\frac{1}{\sqrt{k}}\leq \lim_{n\rightarrow \infty}2\frac{\left(\sqrt{n}\right)}{\sqrt{n}}$$
So $$\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\sum^{n}_{k=1}\frac{1}{\sqrt{k}} = 2$$
