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Hello I am trying to prove the following series diverges $$\sum_{k=1}^{\infty} \ln\left(1+\frac{(-1)^{k+1}}{\sqrt{k+1}}\right)$$ This series alternates around 0 and goes to zero but fails the alternating test as it is not decreasing. Can someone give me a hint as to what test is appropriate here?

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  • $\begingroup$ Hint: To get a good understanding of the behaviour of the series, add the terms with $k=2t+1$ and with $k=2t+2$. The sum of the two logarithms is the logarithm of $\left(1+\frac{1}{\sqrt{2t+2}}\right)\left(1-\frac{1}{\sqrt{2t+3}}\right)$. Expand. Because of cancellation, the middle term is "small," so in the long run irrelevant. $\endgroup$ – André Nicolas Feb 1 '15 at 21:37
  • $\begingroup$ Ok so I got it down to $$\sum_{t=0}^{\infty} ln(1+\frac{\sqrt{2t+3}-\sqrt{2t+2}-1}{\sqrt{2t+2}\sqrt{2t+3}})$$ you are saying I can ignore first tow parts of upper fraction. Then its just a series coming up to 0 from negatives? $\endgroup$ – Rummi Feb 1 '15 at 21:57
  • $\begingroup$ Multiply out. I prefer to leave the product as $1+\frac{\sqrt{2t+3}-\sqrt{2t+2}}{\sqrt{2t+3}\sqrt{2t+2}}-\frac{1}{\sqrt{2t+3}\sqrt{2t+2}}$. The middle term (rationalize the numerator) behaves like $\frac{c}{t^{3/2}}$ for some constant $c$, so in particular is after a while smaller than $\frac{1}{2\sqrt{2t+3}\sqrt{2t+2}}$. Thus our product after a while is less than $1-\frac{1}{2\sqrt{2t+3}\sqrt{2t+2}}$. So the log is less than $-\frac{1}{2\sqrt{2t+3}\sqrt{2t+2}}$, and now the divergence of the harmonic series does the rest. $\endgroup$ – André Nicolas Feb 1 '15 at 22:21
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Since, by the Taylor expansion, as $x \to 0$, you have $$ \ln (1+x)=x-\frac12x^2+\mathcal{O}(x^3) $$ Then you may write, for $k \to +\infty$, $$ \ln\left(1+\frac{(-1)^{k+1}}{\sqrt{k+1}}\right)=\frac{(-1)^{k+1}}{\sqrt{k+1}}-\frac{1}{2(k+1)}+\mathcal{O}\left(\frac{1}{(k+1)^{3/2}}\right) $$ giving $$ \sum\ln\left(1+\frac{(-1)^{k+1}}{\sqrt{k+1}}\right)=\sum \frac{(-1)^{k+1}}{\sqrt{k+1}}-\sum \frac{1}{2(k+1)}+\sum\mathcal{O}\left(\frac{1}{(k+1)^{3/2}}\right), $$ on the right hand side, the first and the last series are convergent, the second (harmonic) series is divergent, thus your initial series is divergent.

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  • $\begingroup$ How do you know the last series is convergent? $\endgroup$ – Rummi Feb 1 '15 at 22:47
  • $\begingroup$ The P-series test $\endgroup$ – dalastboss Feb 1 '15 at 23:07
  • $\begingroup$ but theirs a function there? $\endgroup$ – Rummi Feb 1 '15 at 23:14
  • $\begingroup$ @Rummi There exists a constant $C$ such that, as $k \to +\infty$, $$\left|O\left(\frac{1}{(k+1)^{3/2}}\right)\right| \leq \frac{C}{(k+1)^{3/2}}$$ giving $$\left|\sum O\left(\frac{1}{(k+1)^{3/2}}\right)\right| \leq \sum \frac{C}{(k+1)^{3/2}}$$ the latter series being convergent since $3/2>1$. $\endgroup$ – Olivier Oloa Feb 2 '15 at 5:10

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