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Please if any one can find the following limit:

Let $b$ be a real number, $b>1$. Compute $$ \lim_{n\to\infty}n\cdot\biggl(\frac1{b^n}+\frac1{b^{2n}}+\frac1{b^{3n}} +\dots+\frac1{b^{(n-1)n}}+\frac1{b^{n^2}}\biggr). $$

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Let's squeeze! $$ 0 \leq n\left(\frac{1}{b^n} + \dots + \frac{1}{b^{n^2}}\right) \leq n\times \frac{n}{b^n} \xrightarrow[n\to\infty]{} 0 $$

Edited according to Matt's comment.

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Well, that is mean that the limit is 0 right? but could you please tell how n\times \frac{n^2}{b^n} \xrightarrow[n\to\infty]{} 0 – LoveMath Apr 12 '13 at 8:46
@Ju'x: shouldn't there be only $n^2$ on the right hand side of the inequality (instead of $n^3$)? There are only $n$ terms between the parentheses. – Matt L. Apr 12 '13 at 8:49
@user50382: check this… – Siméon Apr 12 '13 at 8:50
@Matt: of course you're right, but I needed not be very sharp. – Siméon Apr 12 '13 at 8:51

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