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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?

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2 Answers

up vote 9 down vote accepted

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 $$e^x$$ ? – Julian Assange Dec 10 '12 at 22:04
Ah, we solve x later on and then take the logarithm of that? – Julian Assange Dec 10 '12 at 22:09
1  
@JulianAssange $x = \ln(a)$ – Tom Oldfield Dec 10 '12 at 22:25
up vote 0 down vote

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. – GEdgar Dec 10 '12 at 22:15

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