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I have the following to prove:

$1^{k} + 2^{k} + \cdots + n^{k} \text{ is }O (n^{k+1})$

I have done the following:

$$\frac {1^{k} + 2^{k} + \cdots + n^{k}}{n^k} \leq n$$

Am I on the right track? I don't know how to proceed further and would appreciate some feedback/help.

Thanks!

E W H Lee
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Tyrion
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3 Answers3

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$$0\leqslant1^k+2^k+\cdots+n^k\leqslant n^k+n^k+\cdots+n^k=n^{k+1}\in O(n^{k+1})$$

Edit:

Exercise: Use the same idea to show the stronger result that $1^k+2^k+\cdots+n^k\in\Theta(n^{k+1})$.

A starting point could be to note that, for every $i$ between $\frac12n$ and $n$, $i^k\geqslant\frac1{2^k}n^k$. There are $\frac12n$ such terms hence $$1^k+2^k+\cdots+n^k\geqslant\frac12n\cdot\frac1{2^k}n^k=\frac1{2^{k+1}}n^{k+1}.$$ Naturally, all this is suboptimal as far as explicit constants are concerned but it fully suffices to show that, for every fixed nonnegative $k$, $$1^k+2^k+\cdots+n^k\in\Theta(n^{k+1}).$$

Did
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  • I'm not sure If I follow. Btw the question says let k be a fixed constant, am I allowed to set k = 1 and using that fixed constant prove it? – Tyrion Sep 28 '14 at 17:35
  • Here $k$ is an arbitrary constant, you're not allowed to pick what it is. Still, the above is already a proof that it is in $O(n^{k+1})$. It's not, by itself, a proof that it is in $\Theta(n^{k+1})$, although this is also true. – Ian Sep 28 '14 at 17:36
  • I just don't get the right hand side where you started summing nk+nk...etc. The function isn't a summation to begin with so I don't know why you are able to just do that. – Tyrion Sep 28 '14 at 17:39
  • Each $i^k$ for $1\leqslant i\leqslant n$ is bounded by $n^k$ and there are $n$ of them. – Did Sep 28 '14 at 17:49
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Hint Use Stolz Cezaro for $$a_n=\frac{1^k+2^k+..+n^k}{n^{k+1}}$$

N. S.
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Since $1^k+2^k\dots+n^k=\sum_{j=1}^{k+1} {k+1\brace j}\binom{n}{j}(j-1)!$ and this is a polynomial over $n$ of degree $k+1$ we are done.

Asinomás
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