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I am trying to determine whether this series converges or diverges: $\sum_{n=1}^{\infty }\left(\frac{2n+5}{7n+6}\right)^{n\log(n+1)}$.

Here is my solution: I called: $a_{n}=\left(\frac{2n+5}{7n+6}\right)^{n\log(n+1)}$. Then, I used the root test as follows: $\lim_{n \to \infty }\left | a_{n} \right |^{\frac{1}{n}}=\lim_{n \to \infty}\left(\frac{2n+5}{7n+6}\right)^{\log(n+1)}$. Then I called $x_{n}=(\frac{2n+5}{7n+6})^{\log(n+1)}$, Instead of computing $\lim_{n \to \infty}x_{n}$, I computed first $$\lim_{n \to \infty}\log(x_{n})=\lim_{n \to \infty}\log(n)\log\left(\frac{2n+5}{7n+6}\right)=\log(\frac{2}{7})\lim_{n \to \infty}\log(n)=-\infty,$$ therefore: $\lim_{n \to \infty}x_{n}=\lim_{n \to \infty}e^{\log(x_{n})}=0$. Therefore: $\lim_{n \to \infty }\left | a_{n} \right |^{\frac{1}{n}}=0< 1$. So by the root test, the series converges.

Can you please let me know whether my solution is correct (especially the last steps) or not? if there is a mistake, please let me know how I should fix it. Also, if you are aware of a better way of solving the problem , please do let me know. Thanks!

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    $\begingroup$ Correct. The root test will be easiest to apply. Another way is, comparison test with $C(2/7)^n$. $\endgroup$ – Sungjin Kim Jun 19 '13 at 20:09
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    $\begingroup$ One could note $\bigl({2n+5\over 7n+6}\bigr)^{n\log(n+1)}\le\bigl({3n\over 7n} \bigr)^{n\log(n+1)}\le (3/7)^n$ for $n>4$, and then apply the Comparison test. $\endgroup$ – David Mitra Jun 19 '13 at 20:43
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Another way to to prove the convergence would be to study the integral

$$\int_0^\infty \left(\frac{2x+5}{7x+6} \right)^{x\ln x}\, dx.$$

The integrand is a continuous function in $x\to 0$ and on infinity is majorated by $c(2/7)^x$ for some positive constant $c$, hence the integral converges; by an integral criterion, so does the series.

Your method works, too.

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A bit more complicated than I thought:

$$\left (\frac{2 n+5}{7 n+6}\right)^{n \log{(n+1)}} \sim \left (\frac{2 }{7}\right)^{n \log{n}} \frac{\left(1+\frac{5/2}{n}\right)^{n \log{n}}}{\left(1+\frac{6/7}{n}\right)^{n \log{n}}}$$

This becomes

$$\left (\frac{2 }{7}\right)^{n \log{n}} n^{23/14} \quad (n \to \infty)$$

So then the question becomes, does

$$\sum_{n=1}^{\infty} \left (\frac{2 }{7}\right)^{n \log{n} } n^{23/14}$$

converge? The answer is yes, by the integral test, because

$$\int_1^{\infty} dt \, t^{\alpha} \, e^{- \beta t}$$

converges for any positive $\alpha$ and $\beta$, and because $n \lt n \log{n}$. (Here, $\alpha = 23/14$ and $\beta = \log{(7/2)}$.)

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    $\begingroup$ I don't know if the end result is right or not, but you need much more justification than that. For example $\left( \frac{n+1}{n} \right)^n \to e$, but $\left(\frac{n}{n}\right)^n = 1 \to 1$, so clearly the two are not equivalent, even though $\frac{n+1}{n} \sim \frac{n}{n}$. In general, $a_n \sim b_n$ does not imply $a_n^{n \log n} \sim b_n^{n \log n}$ $\endgroup$ – Najib Idrissi Jun 19 '13 at 20:08
  • $\begingroup$ @nik: you were right. Thanks. $\endgroup$ – Ron Gordon Jun 19 '13 at 20:23
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As for $n \geq 3$ we have $$\frac{2n+5}{7n+6} < \frac{3}{7} <1$$

we get for $n \geq 3$

$$\left(\frac{2n+5}{7n+6}\right)^{n\log(n+1)} < \left(\frac{3}{7}\right)^{n\log(n+1)}< \left(\frac{3}{7}\right)^{n}$$

Thus, by comparison to a geometric series, the series is convergent.

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