Harmonic series variant solution

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):=mn=1cosn

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

mn=1cosnn=C(m)m+m1n=1C(n)(1n1n+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)=mn=1ein+ein2=ei2(eim1ei1)+ei2(eim1ei1)=(eim1)(1ei)+(eim1)(1ei)2(ei1)(ei1)=(eim+eim)(ei(m+1)+ei(m+1))2(2(ei+ei))12=cosmcos(m+1)2(1cos1)12

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

|C(m)|11cos1+12<3.

From this it follows that C(m)/m0 as m, and after taking the limit (1) becomes

n=1cosnn=n=1C(n)(1n1n+1),

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

1n1n+1=1n2+n1n2

we have |C(n)(1n1n+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.