For an integer \(a>1\), can \(\sum _{ n=1 }^{ k }{ \sqrt [ a ]{ n } } \) ever be an integer for any integer \(k>1\)?

For \(a=2\), the answer is most likely no. However, I don't have one single clue of how to prove this with \(a=2\), let alone the generalization!

Is there a definitive answer for all integer \(a\)? If so, how?

(It might have something to do with Calculus here, it looks like some kind of series...)

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestIt doesn't make sense to me, do you mean the n to be an x? Otherwise, you are just multiplying the root by the value of k

Log in to reply

Thanks for your info. I didn't have much time when I was typing down.

Any ideas?

Log in to reply

I agree that it is probably no, but not sure how to prove it

Log in to reply

Log in to reply

While doing some research about this, I came across this blog post from almost a decade ago, which applies to your problem for \(a = 2.\) As for higher values of \(a,\) I haven't found anything yet, but perhaps the methods in the blog post can help.

Log in to reply

Yes I believe this result implies the problem for \(a=2\) can never yield an integer. If your sum is \(\sum_{k=1}^n \sqrt{k} =N \), then setting \(s_1=1-N, r_1=1,\) and \(k=r_ks_k^2\) for squarefree \(r_k\), \(2 \leq k \leq n\), then \(\sum_{k=1}^n s_k \sqrt{r_k} =0 \), contradicting the result in the blog.

Log in to reply

Thanks! I’ll check it out.

Log in to reply

this is for series i think

Log in to reply