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Let $b_1=1$ and $b_j=\frac{1}{2j-2}+\frac{1}{2j-1} \forall j\ge2$

(i) Show that the series $\sum_{j=1}^\infty (-1)^{j+1}b_j$ converges.

For this part, given that it's an alternating series, I would think I would have to use the Alternating Series Test.

I'm first going to split it up: $$\sum_{j=1}^\infty (-1)^{j+1}b_j=(-1)^2b_1+\sum_{j=2}^\infty (-1)^{j+1}b_j=1+\sum_{j=2}^\infty (-1)^{j+1}b_j$$ So it remains to show the test works on $b_j$ for $j\ge 2$

It's clear that $b_j \ge 0 \space$ for $j\ge 2$ (I won't bore you with effortless explanations)

How i have to show that $b_j \ge b_{j+1}$ (i.e. decreasing) $$2j\ge 2j-2 \implies \frac{1}{2j} \le \frac{1}{2j-2}$$ $$2j+1 \ge 2j-1 \implies \frac{1}{2j+1} \le \frac{1}{2j-1}$$ Putting them together i get: $$b_j=\frac{1}{2j-2}+\frac{1}{2j-1}\ge \frac{1}{2j}+\frac{1}{2j-1} \ge \frac{1}{2j}+\frac{1}{2j+1}=b_{j+1}$$

Again, it is clear that $\lim b_j=0$ for $j\ge 2$. $$\lim b_j=\lim \frac{\frac{1}{j}}{2-\frac{2}{j}}+\frac{\frac{1}{j}}{2-\frac{1}{j}}=\frac{0}{2-0}+\frac{0}{2-0}=0$$

My question here is can i do it this way? By splitting the summation up.

For the next 2 parts I have no idea what to do. Should i do induction? Should i explicitly calculate it out? Any hints could be appreciated.

(ii) Let $(s_n)$ be the sequence of partial sums of the series $$\sum_{m=1}^\infty\frac{(-1)^{[\frac{m}{2}]}}{m}=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-...$$ Here $[a]$ denotes the greatest integer less than or equal to $a$. Show that for each $k \in \mathbb N$, $s_{2k-1}$ is the $k$-th partial sum of the series $\sum_{j=1}^\infty(-1)^{j+1}b_j$.

(iii) Show that $\lim_{k\to \infty}s_{2k}=\lim_{k\to \infty}s_{2k-1}$. Deduce that the series $\sum_{m=1}^\infty\frac{(-1)^\frac{m}{2}}{m}$ is convergent.

Danxe
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  • First, you are using the phrase "vacuously clear" in a wrong way. Second, if you have fully completed your analysis for the part (i), I don't understand what is the problem for parts (ii) and (iii). Can you show that $\displaystyle{\sum_{n=1}^{\infty} \dfrac{(-1)^{\lfloor \frac{m}{2}\rfloor}}{m}=\sum_{j=1}^{\infty} (-1)^{j+1}b_j}$? – Darío G Mar 20 '16 at 07:23
  • $\sum\limits_{m=1}^{2k-1}\frac{(-1)^{\left \lceil{\frac{m}{2}}\right \rceil }}{m} = 1+\sum\limits_{m=2}^k (-1)^{m+1}(\frac{1}{2m-2}+\frac{1}{2m-1})$. you can change the order of the terms in there freely when you consider the partial sums. – dezdichado Mar 20 '16 at 07:27
  • @Wore Ok I've adjusted the question. That actually the issue i'm having. I can't see the link between the 2 sums. – Danxe Mar 20 '16 at 07:38
  • @dezdichado Hmm... I can't see how your derive that equality. Thanks for the response btw – Danxe Mar 20 '16 at 07:39
  • @dezdichado is it just a definition or some sort? – Danxe Mar 20 '16 at 07:59
  • @Danxe if you can't see it directly, you can prove it by induction. Though it is evident if you just look at the sum you wrote $1+(-1)^1(\frac{1}{2}+\frac{1}{3})+(-1)^2(\frac{1}{4}+\frac{1}{5})+...$ – dezdichado Mar 20 '16 at 08:40

1 Answers1

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Let we do everything at once.

$$\sum_{n=0}^{N}\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)(-1)^n = \sum_{n=0}^{N}(-1)^n \int_{0}^{1} x^{2n}(1-x)\,dx \tag{1}$$ hence: $$ \sum_{n\geq 0}\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)(-1)^n = \int_{0}^{1}\frac{1-x}{1+x^2}\,dx =\color{red}{\frac{\pi}{4}-\frac{\log 2}{2}}.\tag{2}$$ The convergence of the RHS of $(2)$ also follows from Dirichlet's test, since $\{(-1)^n\}_{n\geq 0}$ has bounded partial sums and $\frac{1}{(2n+1)(2n+2)}$ is decreasing to zero.

Jack D'Aurizio
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