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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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    What is Srbin (pozdrav)?2012-09-11
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    @Peter: I suspect that as well, but I'm trying to give the poster the benefit of the doubt.2012-09-11
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    Pozdrav means "greetings" in Czech, and "srbin" means "a Serb".2012-09-11
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    If you're just interested in how the number $1^5 + 2^5 + \cdots (10^n)^5$ begins in base $10$, then it may make more sense to investigate $(1^5 + 2^5 + \cdots + (10^n)^5)/10^{5n}$ (or something).2012-09-11
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    Marko, your observation follows from a general fact $$1^m+2^m+\cdots+n^m \approx \frac{1}{m+1}n^{m+1}.$$2012-09-11
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    It still makes no sense that the sum of positive numbers larger than $1/6$ can be $1/6$.2012-09-11
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    @MichaelGreinecker I think there is some regularization going on. Maybe of if the OP explained.2012-09-11
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    what the heck is this OP I am seeing of late in all the comments?2012-09-11
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    @ajay "OP" is an abbreviation for "original poster". In this case the original poster is marko.2012-09-11
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    @MJD thanks for enlightening me :) I am still wondering if this question makes any sense at all.2012-09-11
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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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