Does $\sum_{n=1}^{\infty}\frac{1}{\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n}}$ converge? Looking for exercises for my students, I stumbled upon this series
$$\sum_{n=1}^{\infty}(-1)^n\frac{1}{\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n}}$$
which can be easily seen to be convergent thanks to the Leibniz Criterion. I forgot tho to put the $(-1)^n$ in the equation and started to try to solve it! Eventually I found out I missed a piece. Anyway I convinced myself that
$$\sum_{n=1}^{\infty}\frac{1}{\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n}}$$ should converge anyway, but am unable to prove it!
Thanks!
 A: Note that
$$
\newcommand\diff{\,\mathrm d}
\sqrt k = \int_{k-1}^k \sqrt k \diff x \geqslant \int_{k-1}^k \sqrt x \diff x, 
$$
then
$$
\sum_1^n \sqrt k \geqslant \sum_1^n \int_{k-1}^k \sqrt x \diff x = \int_0^n \sqrt x \diff x = \frac 23 x^{3/2}\bigg\vert_0^n = \frac 23 n^{3/2},
$$
hence the convergence.
UPDATE
If integrals are not allowed to use, then assume we can use the MVT, note that
$$
\frac 23 (k+1)^{3/2} - \frac 23 k^{3/2} = \left.\frac\diff {\diff x} \left(\frac 23 x^{3/2}\right)\right|_{x=k+t} (k+1 -k) = \sqrt {k+t} \leqslant \sqrt {k+1}
$$
where $t \in (0,1)$, thus
$$
\sum_0^{n-1}\sqrt {k+1} \geqslant \frac 23 n^{3/2}.
$$
This seems different, but this is just a formal change. This is the reverse usage of MVT, which would be natural after learning integrals.
A: You can rewrite the general term as
$$
\frac{1}{\sqrt{1}+\sqrt{2}+\dots+\sqrt{n}} = \frac{1}{n}\frac{n}{\sqrt{1}+\sqrt{2}+\dots+\sqrt{n}} \le \frac{1}{n}\frac{1}{(n!)^{1/2n}},
$$
by the arithmetic-geometric means inequality. By writing
$$
(n!)^2 = [n\cdot 1]\cdot[(n-1)\cdot 2]\cdot\dots\cdot[1\cdot n] \ge n\cdot n\cdot \dots \cdot n = n^n,
$$
and taking $4n$-th roots, we obtain $(n!)^{1/2n} \ge n^{1/4}$. Hence
$$
\frac{1}{n}\frac{1}{(n!)^{1/2n}} \le \frac{1}{n}\frac{1}{n^{1/4}} = \frac{1}{n^{5/4}},
$$
and
$$
\sum_{n=1}^\infty \frac{1}{n^{5/4}}
$$
converges by the $p$-test. Hence the original series converges by comparison.
A: A geometric way to see that the sum converges.
For each $n\ge 1$, $\sqrt{n}\ge\sqrt{x}$ for $x\in [n-1,n]$.
Thus, the denominator $\sqrt{1} + \sqrt{2} + \dots + \sqrt{n}$ is greater than the area $A_n$ below the graph of $\sqrt x$ from $x = 0$ to $x = n$, as can be seen by dividing the interval $[0,n]$ into $n$ subintervals $[0,1],[1,2],\dots,[n-1,n]$ and comparing $\sqrt{n}$ to $\sqrt{x}$ on each subinterval. 
Next, the area $A_n$ below the graph of $\sqrt{x}$ from $x = 0$ to $x = n$ is also greater than the area of the triangle $T_n$ with vertices $(0,0),(n,0),(n,\sqrt{n})$ in the plane, which has area
$$
T_n = \frac{n\sqrt{n}}{2} = \frac{n^{3/2}}{2}.
$$
In sum, we have
$$
\sqrt{1} + \sqrt{2} + \dots + \sqrt{n} \ge A_n \ge T_n = \frac{n^{3/2}}{2},
$$
so by comparison,
$$
\sum_{n=1}^\infty\frac{1}{\sqrt 1 + \sqrt 2 + \dots + \sqrt n} \le \sum_{n=1}^\infty\frac{2}{n^{3/2}},
$$
which converges by the $p$-test.

To compare the estimates I used in my two answers numerically:
By using the arithmetic-geometric means inequality, Wolfram gives
$$
\sum_{n=1}^\infty\frac{1}{n^{5/4}} \approx 4.595,
$$
and by using the geometric method:
$$
\sum_{n=1}^\infty\frac{2}{n^{3/2}} \approx 5.225,
$$
so the AM-GM inequality gives a slightly closer estimate. However, Wolfram approximates the value of the original series as
$$
\sum_{n=1}^\infty\frac{1}{\sqrt 1+\sqrt 2+ \dots +\sqrt n} \approx 2.956,
$$
so neither upper bound is very good.
A: Claim:
$$h(n)=:\frac{2}{3}n\sqrt{n}+\frac{1}{3}\sqrt{n}\leq f(n):=\sum_{i=1}^n\sqrt{i}\leq\frac{2}{3}n\sqrt{n}+\frac{1}{2}\sqrt{n}=g(n)$$
Enough to prove
$$
g(1)\geq f(1)\geq h(1)\wedge g(n)-g(n-1)\geq f(n)-f(n-1)\geq h(n)-h(n-1)
$$
1st part: $g(n)-g(n-1)\geq f(n)-f(n-1)$
$$
\frac{2(n\sqrt{n}-(n-1)\sqrt{n-1})}{3}+\frac{\sqrt{n}-\sqrt{n-1}}{2}\geq\sqrt{n}
$$
Multiply by $6$ and reorganize:
$$
(4n+3-6)\sqrt{n}=(4n-3)\sqrt{n}\geq(4n-1)\sqrt{n-1}=(4n-4+3)\sqrt{n-1}
$$
Squaring both sides and expanding yields:
$$
16n^3-24n^2+9n\geq (16n^2-8n-1)(n-1)=16 n^3-24 n^2+9 n-1
$$
2nd part: $ f(n)-f(n-1)\geq h(n)-h(n-1)$
$$
\frac{2}{3}n\sqrt{n}+\frac{1}{3}\sqrt{n}-\frac{2}{3}(n-1)\sqrt{n-1}-\frac{1}{3}\sqrt{n-1}\leq\sqrt{n}
$$
Multiply by $3$ and reorganize:
$$
(2n-2)\sqrt{n}\leq (2n-1)\sqrt{(n-1)}
$$
$$
4 n^3-8 n^2+4 n\leq 4 n^3-8 n^2+5 n-1
$$
Which is true, hence the claim. Finally, as all the terms are positive:
$$
\sum_{j=1}^{N}\frac{1}{h(j)}\geq\sum_{j=1}^{N}\frac{1}{f(j)}\geq\sum_{j=1}^{N}\frac{1}{g(j)}
$$
A: If you enjoy generalized harmonic numbers
$$\sum_{i=1}^n \sqrt i=H_n^{\left(-\frac{1}{2}\right)}$$ uSe the asymptotics
$$H_n^{\left(-\frac{1}{2}\right)}=\frac{2 n^{3/2}}{3}+\frac{\sqrt{n}}{2}+\zeta
   \left(-\frac{1}{2}\right)+\frac{1}{24\sqrt n}+O\left(\frac{1}{n^{5/2}}
   \right)$$
$$\frac 1 {H_n^{\left(-\frac{1}{2}\right)}}=\frac{3}{2} \frac{1}{n^{3/2}}-\frac{9}{8}\frac{1}{n^{5/2}}
   -\frac{9 \zeta \left(-\frac{1}{2}\right)}{4
   n^3}+O\left(\frac{1}{n^{7/2}}\right)$$
$$\sum_{n=1}^\infty \frac 1 {H_n^{\left(-\frac{1}{2}\right)}}=\frac{3 }{2}\zeta \left(\frac{3}{2}\right)-\frac{9 }{8}\zeta \left(\frac{5}{2}\right)-\frac 94\zeta \left(-\frac{1}{2}\right)\zeta (3)+\cdots$$ Truncated to this level, the summation is approximated by $2.97164$ while the infinite sum would be $\approx 3.16783$.
