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assume $$x_n=\frac{n+1}{2^{n+1}}\sum_{k=1}^n\frac{2^k}{k} ,n=1,2,.....$$

how compute $\lim_{n\to +\infty}x_n$?

Thanks for any hint

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If you are aware of the Cesaro-Stolz theorem, the answer is straightforward. –  sos440 Feb 26 '13 at 19:11
    
@Maisam, I think you want the limit when $\,n\to\infty\,$... –  DonAntonio Feb 26 '13 at 19:42
    
A crude approach that should work is to bound the sum by integrals, and estimate the integrals using integration by parts. –  André Nicolas Feb 26 '13 at 19:47

1 Answer 1

up vote 4 down vote accepted

Since the answer is straightforward if we exploit the Cesaro-Stolz theorem, here I want to present a solution that does not use this theorem.

Notice first that whenever $1<r<2$, we have

$$ \lim_{n\to\infty} \frac{\sum_{k=1}^{n} \frac{r^k}{k}}{\frac{2^{n+1}}{n+1}} = 0.$$

Indeed, the numerator is dominated by $n r^n$, thus we have

$$ 0 \leq \frac{\sum_{k=1}^{n} \frac{r^k}{k}}{\frac{2^{n+1}}{n+1}} \leq \frac{n(n+1)}{2} \left(\frac{r}{2}\right)^n \xrightarrow[]{n\to\infty} 0.$$

This shows that, the limit in question exists with value $\ell$ if and only if

$$ \lim_{n\to\infty} \frac{\sum_{k=1}^{n} \frac{2^k - r^k}{k}}{\frac{2^{n+1}}{n+1}} = \ell.$$

But note that

$$ \sum_{k=1}^{n} \frac{2^k - r^k}{k} = \sum_{k=1}^{n} \int_r^2 x^{k-1} \; dx = \int_r^2 \frac{x^{n} - 1}{x-1} \; dx. $$

This in particular shows that

$$ \sum_{k=1}^{n} \frac{2^k - r^k}{k} \leq \int_r^2 \frac{x^{n}-1}{r-1} \; dx \leq \frac{2^{n+1}-r^{n+1}-(n+1)(2-r)}{(r-1)(n+1)} $$

and

$$ \sum_{k=1}^{n} \frac{2^k - r^k}{k} \geq \int_r^2 (x^{n}-1) \; dx \geq \frac{2^{n+1}-r^{n+1}-(n+1)(2-r)}{n+1} $$

These equalities show that

$$ 1\leq \liminf_{n\to\infty} \frac{\sum_{k=1}^{n} \frac{2^k - r^k}{k}}{\frac{2^{n+1}}{n+1}} \leq \limsup_{n\to\infty} \frac{\sum_{k=1}^{n} \frac{2^k - r^k}{k}}{\frac{2^{n+1}}{n+1}} \leq \frac{1}{r-1}. $$

Now since $r$ was arbitrary, taking $r\to2^{-}$ yields the limit $\ell = 1$.

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