infinity sum of numbers 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.

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.
