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$$1 + 2^x + 3^x + 4^x + ... + n^x + ...$$

I'm not sure how to approach this one? I should be able to show this with just the basic library of Calc II tests (Geometric, p-series, Divergence (nth term) test, Integral test, Direct Comparison test, Alternating Series Test, or Ratio test)

$$\sum_{n=1}^\infty n^x = \sum_{n=1}^\infty \frac{1}{n^{-x}}$$

Is this now a p-series?

EDIT:
Well, p-series converges when p>1.
So, I need $-x > 1$

Which means $x < -1$

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closed as off-topic by heropup, lulu, Michael Hoppe, Shailesh, choco_addicted Mar 25 '16 at 1:25

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  • $\begingroup$ This site is not a homework answering service, which is clearly suggested by your posting pattern. Stop abusing it. $\endgroup$ – heropup Mar 24 '16 at 21:34
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    $\begingroup$ Can I downvote your answer? I'm not a student. I'm an adult learning Calculus independently. $\endgroup$ – JackOfAll Mar 24 '16 at 22:06
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    $\begingroup$ No, you can't downvote me, because I did not answer your question. Regardless, your posting behavior is not consistent with site guidelines. Questions posted to this site generally require you to provide context, which includes your own efforts to solve the question. $\endgroup$ – heropup Mar 24 '16 at 22:15
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    $\begingroup$ -1 for threatening to downvote someone's answer for pointing out the proper use of the site. (Not actually downvoting the question, but I want to express my displeasure with OP.) $\endgroup$ – Brian Tung Mar 24 '16 at 22:21
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I don't know what you call a p-serie but $\sum \frac{1}{n^p}$ converges if and only if $p>1$ so $-x>1$ and finally $x<-1$

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  • $\begingroup$ Ugh! I had the rule backwards. Converges if p>1. Thanks!!! $\endgroup$ – JackOfAll Mar 24 '16 at 22:19
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We see that

$$\sum_{n=1}^{\infty}n^x\approx\int_1^{\infty}n^xdn$$

$$=\lim_{n\to\infty}\frac1{x+1}n^{x+1}-\frac1{x+1}1^{x+1}$$

So, this converges if $\lim_{n\to\infty}n^{x+1}$ converges, implying $x+1<0$, or $x<-1$

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