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If we have a series of numbers $1^5 + 2^5 + 3^5 + \cdots + (10^n)^5.$ Final sum of the series is approximately equal $16666\ldots$ .

If there is more and more numbers in the series is the result of closer and closer to $16666\ldots$ .

For example if the last number $1000$ or $10000$ or $100000$ and so on, the final sum is closer to $16666\ldots$ . If it is true (of course it is), can we conclude that $1^5 + 2^5 + 3^5 + \cdots = \frac 1 6$

Greetings.

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    @ajay It turns out that [a search for "OP" on Meta](http://meta.math.stackexchange.com/search?q=OP) produces [What does "OP" mean?](http://meta.math.stackexchange.com/questions/4090/what-does-op-mean) as the #1 hit. This suggests a strategy that might be useful if you become similarly puzzled in the future.2012-09-11

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The orthogonal projection of this question onto the subspace of sensible questions is answered by @sos440's comment.

Claim: $\frac{1^m + 2^m + \cdots + n^m}{n^{m+1}}\to \frac{1}{m+1}.$

Proof: Note $\text{LHS} = \frac{1}{n}((1/n)^m + (2/n)^m + \cdots + ((n-1)/n)^m + 1^m)$ is a Riemann sum approximating the integral $\int_0^1 x^m\,dx = \frac{1}{m+1}$.

Thus we can say that $1^5 + 2^5 + \cdots + (10^n)^5 \sim \frac{1}{6} (10^n)^6,$which will look like $1666\ldots$ in base $10$, as you correctly observe.

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    I upvoted this as soon as I read the first sentence, but on later perusal, the rest of the answer was pretty good too.2012-09-11