Convergence of $\sum_{n=1}^\infty\frac{1}{n} [1+\frac{1}{\sqrt{2}}+.....+\frac{1}{\sqrt{n}}]$ Let $t_n= \frac{1}{n} [1+\frac{1}{\sqrt{2}}+.....+\frac{1}{\sqrt{n}}]$, n=1,2...
Then I am asked whether series$ \sum t_n $ converge or diverge. Also whether sequence $ t_n $ converge to zero or not.
I tried About sequence $t_n$, by Cauchy's 1st theorem, that since 1/$\sqrt{n} $ converges to zero so does t_n. But the seriex part is doubtful to me.?
 A: We can use the fact that
$$
\sum_{i=1}^n\frac1{\sqrt i}=\sum_{i=1}^n\frac1{\sqrt i}\int_i^{i+1}\mathrm dx\ge\sum_{i=1}^n\int_i^{i+1}\frac1{\sqrt{x}}\mathrm dx=\int_1^{n+1}\frac1{\sqrt{x}}\mathrm dx=2\sqrt{n+1}-2.
$$
Similarly, $\sum_{i=1}^ni^{-1/2}\le2\sqrt n-1.$
Hence, $\sum_{i=1}^n i^{-1/2}\sim 2\sqrt n$
as $n\to\infty$ and
$$
\frac1n\sum_{i=1}^n\frac1{\sqrt i}\sim \frac2{\sqrt n}
$$
as $n\to\infty$ ($\sim$ means that the ratio of the two sequences goes to $1$ as $n\to\infty$).
A: Can you find simple lower and upper bounds for $t_n$? If the series of the lower bounds diverges, then the original diverges. If the series of the upper bounds converge, then the original converges since the terms are positive. If neither of these hold, then try to find tighter bounds.
By the way, your splitting into sums is correct in this case, but not in general when terms can be negative.
As for $t_n$ itself, you can find an approximate anti-difference as follows:
$\sqrt{n+1} - \sqrt{n} < \frac{1}{\sqrt{n}} < \sqrt{n} - \sqrt{n-1}$
You can use these to get lower and upper bounds for $t_n$.
