Last year I asked if the series ∑∞n=1cosn/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):=m∑n=1cosn
Now, using summation by parts the series may be rewritten so that
m∑n=1cosnn=C(m)m+m−1∑n=1C(n)(1n−1n+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)=m∑n=1ein+e−in2=ei2(eim−1ei−1)+e−i2(e−im−1e−i−1)=(eim−1)(1−ei)+(e−im−1)(1−e−i)2(ei−1)(e−i−1)=(eim+e−im)−(ei(m+1)+e−i(m+1))2(2−(ei+e−i))−12=cosm−cos(m+1)2(1−cos1)−12
By the triangle inequality, this has an absolute upper bound of
|C(m)|≤11−cos1+12<3.
From this it follows that C(m)/m→0 as m→∞, and after taking the limit (1) becomes
∞∑n=1cosnn=∞∑n=1C(n)(1n−1n+1),
and we can use the absolute comparison test on the right sum. Using the fact that
1n−1n+1=1n2+n≤1n2
we have |C(n)(1n−1n+1)|<3/n2, and ∑∞n=13/n2=3ζ(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 ∑∞n=1cosn/n converges.