# infinity sum of numbers [closed]

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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## closed as not a real question by Pedro Tamaroff♦, Bill Cook, Michael Greinecker♦, Jason DeVito, WilliamSep 11 '12 at 18:10

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

What is Srbin (pozdrav)? – Jason DeVito Sep 11 '12 at 17:27
Pozdrav means "greetings" in Czech, and "srbin" means "a Serb". – Pedro Tamaroff Sep 11 '12 at 17:29
Marko, your observation follows from a general fact $$1^m+2^m+\cdots+n^m \approx \frac{1}{m+1}n^{m+1}.$$ – Sangchul Lee Sep 11 '12 at 17:35
It still makes no sense that the sum of positive numbers larger than $1/6$ can be $1/6$. – Michael Greinecker Sep 11 '12 at 17:44
@ajay It turns out that a search for "OP" on Meta produces What does "OP" mean? as the #1 hit. This suggests a strategy that might be useful if you become similarly puzzled in the future. – MJD Sep 11 '12 at 18:08

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.