Harmonic series variant solution

Last year I asked if the series $\sum_{n=1}^\infty\cos n/n$ converges or diverges. Although the series looks rather simple, the standard convergence tests learned in Calculus 1 do not apply directly. However, there is a trick which allows one to rewrite the series into a form where the standard tests can be used.

The trick is to use summation by parts, the discrete analog of integration by parts. Although less well-known, summation by parts does have its uses; for example, it was the method of proving Abel’s summation formula which has come up before.

To simplify the exposition, it is convenient to define a function for the partial sums of the series’ numerators:

\[ C(m) := \sum_{n=1}^m \cos n \]

Now, using summation by parts the series may be rewritten so that

\[ \sum_{n=1}^m\frac{\cos n}{n} = \frac{C(m)}{m} + \sum_{n=1}^{m-1} C(n)\Bigl(\frac{1}{n}-\frac{1}{n+1}\Bigr) . \tag{1} \]

The key observation to make now is that $C(m)$ is bounded. This can be seen by using the complex exponential expression of the cosine, and summing a geometric series to find a closed-form expression for $C(m)$, as follows:

C(m) &= \sum_{n=1}^m\frac{e^{in}+e^{-in}}{2} \\
&= \frac{e^i}{2}\Bigl(\frac{e^{im}-1}{e^i-1}\Bigr)+\frac{e^{-i}}{2}\Bigl(\frac{e^{-im}-1}{e^{-i}-1}\Bigr) \\
&= \frac{(e^{im}-1)(1-e^i)+(e^{-im}-1)(1-e^{-i})}{2(e^i-1)(e^{-i}-1)} \\
&= \frac{(e^{im}+e^{-im})-(e^{i(m+1)}+e^{-i(m+1)})}{2(2-(e^i+e^{-i}))}-\frac{1}{2} \\
&= \frac{\cos m-\cos(m+1)}{2(1-\cos1)}-\frac{1}{2}

By the triangle inequality, this has an absolute upper bound of

\[ \lvert C(m)\rvert \leq \frac{1}{1-\cos1}+\frac{1}{2}<3 . \]

From this it follows that $C(m)/m\to0$ as $m\to\infty$, and after taking the limit (1) becomes

\[ \sum_{n=1}^\infty\frac{\cos n}{n} = \sum_{n=1}^\infty{}C(n)\Bigl(\frac{1}{n}-\frac{1}{n+1}\Bigr) , \tag{2} \]

and we can use the absolute comparison test on the right sum. Using the fact that

\[ \frac{1}{n}-\frac{1}{n+1} = \frac{1}{n^2+n} \leq \frac{1}{n^2} \]

we have $\bigl\lvert C(n)(\frac{1}{n}-\frac{1}{n+1})\bigr\rvert<3/n^2$, and $\sum_{n=1}^\infty 3/n^2=3\zeta(2)$ is absolutely convergent, so by the comparison test the right sum in (2) is absolutely convergent as well. Thus the left side of (2) must converge, so $\sum_{n=1}^\infty\cos n/n$ converges.

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