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I have the following series.

$$(-1)^n \times \ln\Bigg(\frac{8n+5}{7n+3}\Bigg)$$

I tried the root, ratio and integral tests, but am doing something wrong because I am unable to tell if this series converges.

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$\sum a_n$ converges, then $a_n\rightarrow 0$. –  Moron plus plus Jul 20 at 6:55

2 Answers 2

Hint:

Use the nth term test.

First consider the inside fraction. What is $\displaystyle \lim_{n \rightarrow \infty} \Bigg( \frac{8n+5}{7n+3} \Bigg)$? Given this, what is the asymptotic behavior when the natural log is applied?

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Applying this to the absolute value of the series, and seeing that this fraction converges to $\frac{8}{7}$, this series is divergent? –  user165293 Jul 20 at 3:27
    
Indeed. In particular, given some $\epsilon > 0$, the $n^{th}$ term will be within $\epsilon$ distance of either $\ln{\frac{8}{7}}$ or $-\ln{\frac{8}{7}}$ for all sufficiently large $n$. So the sequence doesn't limit to $0$. –  Kaj Hansen Jul 20 at 3:35
    
Oh, I should add that the integral test wouldn't apply here since that test requires the series be strictly monotonic beyond a certain point. –  Kaj Hansen Jul 20 at 3:42
    
Is applying a convergence/divergence test to the input of a function and then working your way outwards a general rule? –  user165293 Jul 20 at 6:34
1  
Oh no, I was implying both conditions needs to hold. As you have shown that one condition had already failed, the other didn't matter as much but shouldn't be neglected for the OP's sake. –  Nameless Jul 21 at 1:22

Considering a large value of $n$, rewrite $$ \ln\Bigg(\frac{8n+5}{7n+3}\Bigg)=\ln(8n+5)-\ln(7n+3)=\ln(8n)+\ln\Big(1+\frac{5}{8n}\Big)-\ln(7n)-\ln\Big(1+\frac{3}{7n}\Big)$$ $$ \ln\Bigg(\frac{8n+5}{7n+3}\Bigg)=\ln\Big(\frac{8}{7}\Big)+\ln\Big(1+\frac{5}{8n}\Big)-\ln\Big(1+\frac{3}{7n}\Big)$$ Now, consider, for small values of $y$, the Taylor series $$\ln(1+y) \simeq \frac{y}{1}-\frac{y^2}{2}+\frac{y^3}{3}$$ replace $y=\frac{5}{8n}$ in the first logarithm and $y=\frac{3}{7n}$ in the second logarithm. You will end with $$ \ln\Bigg(\frac{8n+5}{7n+3}\Bigg)= \log \left(\frac{8}{7}\right)+\frac{11}{56 n}-\frac{649}{6272 n^2}+\frac{29051}{526848 n^3}+O\left(\left(\frac{1}{n}\right)^4\right)$$

Applying the same approach to the most general case, you could find that $$ \ln\Bigg(\frac{an+b}{cn+d}\Bigg)= \log \left(\frac{a}{c}\right)+\frac{\frac{b}{a}-\frac{d}{c}}{n}+\frac{\frac{d^2}{2 c^2}-\frac{b^2}{2 a^2}}{n^2}+\frac{\frac{b^3}{3 a^3}-\frac{d^3}{3 c^3}}{n^3}+O\left(\left(\frac{1}{n}\right)^4\right)$$

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This is clever. –  Kaj Hansen Jul 20 at 6:49
    
@KajHansen. Thank you ! Cheers :) –  Claude Leibovici Jul 20 at 6:50

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