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Let $S=1-1/3+1/5-1/7+\cdots$. As each term in the series is decreasing and tends to $0$, it is known that their sum exists and is finite by alternating series test. And by considering $\int_0^11/(1+x^2)dx$, it is known that $S=\pi/4$.

I am wondering is there a fast way to see that $S\neq 0$? In general, is it true that for $T=\sum_{n=1}^\infty (-1)^n a_n$, where $a_n>0$ are strictly decreasing and tend to $0$, $T\neq 0$?

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    $\begingroup$ If you want $T$ to be a generalization of $S$, you'll want to either start at $n=0$ or make it $\sum_{n=1}^\infty (-1)^{n-1} a_n$. $\endgroup$ – Robert Israel Nov 18 at 20:35
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If a sequence converges to a limit, then every subsequence converges, to the same limit. In the case of an infinite convergent sum, this implies that the partial sums with an even number of terms will converge to the same value. However, those partial sums are of positive numbers such as $1-1/3$, $1/5-1/7$, $1/9-1/11$ etc. so they make up an increasing sequence of positive numbers, and so the limit must be positive.

The same applies to your general case.

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The sum is always $\ge$ each even-numbered partial sum and $\le$ each odd-numbered one. So $S \ge 1 - 1/3 = 2/3 > 0$.

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Assuming that you start from $n=0$. Let $b_n = a_{2n} - a_{2n+1}$. Then the infinite sum of $a_n$ is equal to the infinite sum of $b_n$. But $b_n$ is always positive since $a_n$ is strictly decreasing. So the sum is also greater than zero.

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We have that for any $k$

$$\frac1{2k-1}-\frac1{2k+1}=\frac2{4k^2-1}\ge\frac1{2k^2}$$

therefore for any $N\ge 1$

$$S_{2N}=\sum_{k=1}^{2N} \frac{(-1)^{k+1}}{2k-1}\ge\frac12\sum_{k=1}^N \frac1{(2k-1)^2} \ge \frac12 $$

From the given inequality, taking the limit to infinity, we can also deduce that $S\ge \frac{\pi^2}{16}$.

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    $\begingroup$ The question asks for a "fast" way to determine that the series evaluates to something nonzero. Your answer requires the asker to have memorized the value of a different series (which is, to my knowledge, usually computed using Fourier analysis, which is typically taught after calculus and analysis), hence it seems like a lot more work than simply computing the value of the series, as the original asker has done. As such, I don't think that this actually answers the question asked. $\endgroup$ – Xander Henderson Nov 19 at 14:37
  • $\begingroup$ @XanderHenderson Not exactly, indeed from here $S_{2N}=\sum_{k=1}^{2N} \frac{(-1)^{k+1}}{2k-1}\ge\frac12\sum_{k=1}^N \frac1{(2k-1)^2}\ge \frac12$ we can also conlude without any knowledge for $\sum_{k=1}^\infty \frac1{(2k-1)^2}$. That was my first intention indeed, after I've added the result by a suggestion but it is not necessary. $\endgroup$ – user Nov 19 at 14:39
  • $\begingroup$ @XanderHenderson Note indeed that the argument I used is the same of the others answer but with an explicit evaluation usng partial summation. Nothing new of course but also nothing worst. $\endgroup$ – user Nov 19 at 15:05

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