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For which positive values on a does the series converge?:

$$ \sum _{n=1}^{\infty}na^{\ln(n)}$$

I have tried to rewrite the expression, but that gives me nothing.

Anyone got a clue?

2 Answers 2

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I will guess that your $i$ should be an $n$. If $a$ is positive, we may write it as $e^x$ for some $x$. Then $$ na^{\ln(n)}=n(e^x)^{\ln(n)}=ne^{\ln(n^x)}=n(n^x)=n^{x+1}. $$ Now you've got yourself a $p$-series.

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    Yep, changed that. But why can you change a to $$e^x$$ ?2012-12-10
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    Ah, we solve x later on and then take the logarithm of that?2012-12-10
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    @JulianAssange $x = \ln(a)$2012-12-10
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Hint: Use the root test and see what you get.

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    Too bad, root test fails. I would have written "try the ratio test" but that fails, too.2012-12-10