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I'm new to real analysis and trying to solve a basic questions. I'm asked to prove that the following sequence is monotonically decreasing: $b_{n}=\sum_{k=1}^n\frac{(-1)^{2k}}{2k+1}$. So I need to prove $b_{n} \leq b_{n+1}$ Now I think I should use induction.

basisstep: $b_{1}=\frac{1}{3}$.

induction hypothesis: assume $b_{n+1} \geq b_{n}$.

Proof. Since $b_{n+1}=b_{n}+\frac{(-1)^{2n+2}}{2n+3}=b_{n}+\frac{1}{2n+3}$. And observe that $\frac{1}{2n+3}>0$. So it follows that $b_{n+1}=b_{n}+\frac{1}{2n+3}>b_{n}$.

Is this a valid proof or am I missing something?

Thanks in advance.

UPDATE First of all, thanks for the fast comments. I now realise I misinterpreted the proof. Instead of proving that the summation is monotonically decreasing, I need to prove that the elements consisting of (1/3,1/5,1/7..1/(2n+1) is monotonically decreasing.

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  • $\begingroup$ You need to prove $b_{n+1} \leq b_{n+2}$ $\endgroup$ – user392395 Nov 24 '17 at 15:21
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    $\begingroup$ It seems correct to me. But this sequence is monotonically increasing. $\endgroup$ – Math Lover Nov 24 '17 at 15:21
  • $\begingroup$ what is your $b_n$? $\endgroup$ – Dr. Sonnhard Graubner Nov 24 '17 at 15:21
  • $\begingroup$ i have found it sorry $\endgroup$ – Dr. Sonnhard Graubner Nov 24 '17 at 15:26
  • $\begingroup$ I see what you mean with it being monotonically increasing, yet the question states that I need to prove that it decreases... I see that because it is a summation the total sum increases by an ever smaller 'element'. $\endgroup$ – Mathbeginner Nov 24 '17 at 15:45
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simplify your calculations and observe that $$(-1)^{2k}=1$$ for all $k$

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  • $\begingroup$ good point, thanks $\endgroup$ – Mathbeginner Nov 24 '17 at 15:45

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