Linked Questions
13 questions linked to/from How close can $\sum_{k=1}^n \sqrt{k}$ be to an integer?
73
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answer
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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 ...
- 27.1k
21
votes
7
answers
960
views
How closely can we estimate $\sum_{i=0}^n \sqrt{i}$
By looking at an integral and bounding the error?
- 321
16
votes
4
answers
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 $n\to \infty$?
- 991
25
votes
4
answers
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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 ...
- 8,648
41
votes
2
answers
3k
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 ...
- 39.7k
11
votes
3
answers
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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
1
answer
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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 ...
- 17k
6
votes
2
answers
1k
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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
$$
\sum_{k=1}^n \sqrt k = \frac{2 \sqrt{n^3}}{3} -\frac{2}{3} + \frac 1 ...
- 163
8
votes
1
answer
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$S=\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{4}+\dots+\sqrt{m}$ is almost an integer. Find $m$
For an integers $m$ and $n$, $1<m\le n$ , we need to find the best $m$ so that $S=\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{4}+\dots+\sqrt{m}$ is almost an integer.
Example: when $n=40$, then the best ...
- 4,800
-2
votes
1
answer
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Sum of finite series involving square roots [duplicate]
What's the the result of:
$$\sum_{k=1}^{n}{\sqrt{k}+1}$$
Thanks.
- 155
10
votes
2
answers
239
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 ...
- 12.4k
0
votes
2
answers
91
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
- 499
0
votes
2
answers
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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 ...
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