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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 $0$, $T\neq 0$?
The sum is always $\ge$ each even-numbered partial sum and $\le$ each odd-numbered one. So $S \ge 1 - 1/3 = 2/3 > 0$.
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.
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.
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 $S\ge \frac{\pi^2}{16}$.
