Determine if $\sum_{k=1}^\infty\frac{(-1)^k}{2k-1} \cos(2k-1)$ is a convergent series

Question

$$\sum_{k=1}^\infty\frac{(-1)^k}{2k-1} \cos(2k-1)$$

There, its value $-\pi/4$ is immediately determined. But how to be sure, a priori, that the series would converge?

I am new as regards this topic. I tried the Leibniz test, but it fails: due to the cosine, the series may not have alternating signs, even for large $n$. Then, how to determine if the series is convergent?

@AdityaDwivedi So, you mean using the direct comparison test between the series $\frac{(-1)^k}{2k - 1}$ and $\frac{(-1)^k}{2k - 1} \cos (2k - 1)$? — BowPark, Jun 19 '20 at 08:33
@AdityaDwivedi Thank you. If you want, you can create an answer. — BowPark, Jun 19 '20 at 08:38
@AdityaDwivedi: that test works for. Series of positive terms. — Paramanand Singh, Jun 19 '20 at 08:54
You have accepted a wrong answer. You should try to learn more and fix it. — Paramanand Singh, Jun 19 '20 at 09:26
@ParamanandSingh Yes, this test is only available for series of positive terms. But aren't the series of absolute values $\displaystyle \sum_{k=1}^{\infty} \left| \frac{(-1)^k }{2k-1} \right|$ and $\displaystyle \sum_{k=1}^{\infty} \left| \frac{(-1)^k }{2k-1} \cos(2k - 1) \right|$ series of positive terms? — BowPark, Jun 19 '20 at 09:32
@BowPark Absolute value series $\sum\left|\frac{(-1)^k}{2k-1}\right|$ diverges. — Paramanand Singh, Jun 19 '20 at 09:34
@BowPark: your question is not so trivial to be answered in such a casual manner. In fact the inequalities written in this accepted answer are meaningless. — Paramanand Singh, Jun 19 '20 at 09:35
@ParamanandSingh I see that the ratio test is $1$ as regards $\sum \left| \frac{(-1)^k}{2k - 1} \right|$. If this is immediate (otherwise it would require another question), how do you obtain that it diverges? — BowPark, Jun 19 '20 at 09:38
Well compare the series with $\sum 1/k^p$ with $p=1$ and thus series diverges. Remember harmonic series. — Paramanand Singh, Jun 19 '20 at 09:39

score 1 · Answer 1 · answered Jun 19 '20 at 09:24

Let $$S_n=\sum_{k=1}^{n}(-1)^k\cos(2k-1)$$ and $$T_n=\sum_{k=1}^{n}\cos(2k-1)$$ so that $$S_n+T_n=2\sum_{k\text{ even}, k\leq n} \cos(2k-1)$$ and $$T_n-S_n=2\sum_{k\text{ odd}, k\leq n} \cos(2k-1)$$ Using the formula for sums of trigonometric series with arguments in arithmetic progression one can show that the above sums are bounded for all $n$ so that $S_n$ is also bounded.

Now use $a_n=1/(2n-1)$ so that $a_n$ decreases and tends to $0$ and $b_n=(-1)^{n}\cos(2n-1)$ and the sum $S_n=\sum_{k=1}^{n}b_k$ is bounded and by Dirichlet test the series $\sum a_nb_n$ converges.

Determine if $\sum_{k=1}^\infty\frac{(-1)^k}{2k-1} \cos(2k-1)$ is a convergent series

1 Answers1