Linked Questions

65
votes
1answer
4k views

Is $\sqrt1+\sqrt2+\dots+\sqrt n$ ever an integer?

Related: Can a sum of square roots be an integer? Except for the obvious cases $n=0,1$, are there any values of $n$ such that $\sum_{k=1}^n\sqrt k$ is an integer? How does one even approach such a ...
21
votes
7answers
793 views

How closely can we estimate $\sum_{i=0}^n \sqrt{i}$

By looking at an integral and bounding the error?
16
votes
4answers
1k views

Asymptotic behavior of the partial sums $\sum\limits_{k=1}^{n}k^{1/4} $

What is the asymptotic behavior of the sequence: \begin{equation} s_n=\sum_{k=1}^{n}k^{1/4} \end{equation} when $n\to \infty$?
25
votes
4answers
3k views

Evaluate: $\lim_{n\to\infty}\left({2\sqrt n}-\sum_{k=1}^n\frac1{\sqrt k}\right)$

How to find $\lim\limits_{n\to\infty}\left({2\sqrt n}-\sum\limits_{k=1}^n\frac1{\sqrt k}\right)$ ? And generally does the limit of the integral of $f(x)$ minus the sum of $f(x)$ exist? How to prove ...
34
votes
2answers
2k views

Evaluating the series $\sum_{n=1}^{\infty} \frac{n}{e^{2 \pi n}-1}$ using the inverse Mellin transform

Inspired by this answer, I'm trying to show that $$\sum_{n=1}^{\infty} \frac{n}{e^{2 \pi n}-1} = \frac{1}{24} - \frac{1}{8 \pi}$$ using the inverse Mellin transform. But the answer I get is twice ...
11
votes
3answers
2k views

How to calculate the asymptotic expansion of $\sum \sqrt{k}$?

Denote $u_n:=\sum_{k=1}^n \sqrt{k}$. We can easily see that $$ k^{1/2} = \frac{2}{3} (k^{3/2} - (k-1)^{3/2}) + O(k^{-1/2}),$$ hence $\sum_1^n \sqrt{k} = \frac{2}{3}n^{3/2} + O(n^{1/2})$, because $\...
8
votes
1answer
2k views

Fractional part summation

Let us consider the sum $$\displaystyle S_K=\sum_{n \geq \sqrt{K}}^{2 \sqrt{K}} \left\{ \sqrt {n^2-K} \right\} $$ where $K$ is a positive integer and where $\{ \}$ indicates the fractional part. If ...
6
votes
2answers
678 views

Euler-Maclaurin Summation

Using EM summation formula estimate $$ \sum_{k=1}^n \sqrt k $$ up to the term involving $\frac{1}{\sqrt n}$ My attempt is $$ \sum_{k=1}^n \sqrt k = \frac{2 \sqrt{n^3}}{3} -\frac{2}{3} + \frac 1 ...
8
votes
1answer
305 views

$S=\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{4}+\dots+\sqrt{m}$ is almost an integer. Find $m$

For an integers $m$ and $n$, $1<m\le n$ , we need to find the best $m$ so that $S=\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{4}+\dots+\sqrt{m}$ is almost an integer. Example: when $n=40$, then the best ...
-2
votes
1answer
3k views

Sum of finite series involving square roots [duplicate]

What's the the result of: $$\sum_{k=1}^{n}{\sqrt{k}+1}$$ Thanks.
9
votes
2answers
221 views

Sum of all consecutive natural root differences on a given power

I accidentally observed that $\sqrt{n} - \sqrt{n-1}$ tends to $0$ for higher values of $n$, so I've decided to try to sum it all, but that sum diverged. So I've tried to make it converge by giving it ...
0
votes
2answers
88 views

Help with a limit $S_{n}=\frac{1}{n}\sum _{k=1}^{n} \sqrt{\frac{k}{n}}$

How can I calculate the next limit? $S_{n}=\frac{1}{n}\sum _{k=1}^{n} \sqrt{\frac{k}{n}}$ when "n" goes to infinity
0
votes
2answers
87 views

Does $\lim_{n\to\infty}\left(4\sum_{i=1}^{2n}(-1)^i\sqrt{i}\right)-\sqrt{2n}-\sqrt{2n+1}$ converge?

Ratio test gives me $\displaystyle \lim_{n\to\infty}\frac{\sqrt{2(n+1)}-3\sqrt{1+2(n+1)}+3\sqrt{2+2(n+1)}-\sqrt{3+2(n+1)}}{\sqrt{2n}-3\sqrt{1+2n}+3\sqrt{2+2n}-\sqrt{3+2n}}=1$, I'm not sure how to ...