Convergence of the series $\sum\limits_{n=2}^{\infty} \frac{ (-1)^n} { \ln(n) +\cos(n)}$ $$\sum_{n=2}^{\infty} \frac{ (-1)^n} { \ln(n) +\cos(n)}$$
 A: This replaces my original answer, which was faulty but led in the right direction (I think).
I'll let the series start at some $n_0\ge2$; the reason for this will become apparent later.
Expand in $\cos(n)/\ln(n)$ (with $\lvert\cos(n)/\ln(n)\rvert < 1$ for $n \ge 2$):
\[\sum_{n=n_0}^\infty \sum_{k=0}^\infty \frac{(-1)^n}{\ln(n)} \frac{\cos^k(n)}{\ln^k(n)}\]
Let's interchange the summations and worry later whether that was OK:
\[\sum_{k=0}^\infty \sum_{n=n_0}^\infty \frac{(-1)^n}{\ln(n)} \frac{\cos^k(n)}{\ln^k(n)}\]
Now the inner sum converges according to Dirichlet's test. In fact we can proceed as in the proof of Dirichlet's test to bound its limit (suppressing indices $k$ while we work at fixed $k$):
\[a_n=\ln^{-k-1}(n)\]
\[b_n=(-1)^n\cos^k(n)=\sum_{j=0}^k\alpha_{jk}\cos((j + \pi)n)\;,\]
with $\lvert\alpha_{jk}\rvert \le 1$, using the power-reduction formula for the cosine and absorbing the factor $(-1)^n$ into the cosines by adding $\pi$ to their frequency.
\[B_m=\sum_{n=n_0}^m \; b_n\]
\[\lvert B_m\rvert = \left\lvert\sum_{n=n_0}^m \sum_{j=0}^k
\alpha_{jk}\cos((j + \pi)n)\right\rvert
\le \sum_{j=0}^k\left\lvert\sum_{n=n_0}^m\cos((j + \pi)n)\right\rvert \le \sum_{j=0}^k\frac{2}{1 - \cos(j + \pi)}=:M_k\]
\[\left\lvert\sum_{n=n_0}^m a_n b_n\right\rvert = \left\lvert B_m a_{m+1}+\sum_{n=n_0}^{m}B_n(a_n-a_{n+1})\right\rvert\le\]
\[\le M_k a_{m+1} + \sum_{n=n_0}^{m}M_k(a_n-a_{n+1})=
M_k a_{m+1} + M_k(a_{n_0}-a_{m+1}) = M_ka_{n_0}\]
Thus, we can apply the comparison test with the following series:
\[\sum_{k=0}^\infty M_k\ln^{-k-1}(n_0)\]
This shows why we need to start at some $n_0$ instead of $2$, which wouldn't work since $\ln 2 < 1$.
Now everything depends on the behaviour of $M_k$. The cosine comes arbitrarily close to $1$; when $j$ comes within $\epsilon$ of an odd multiple of $\pi$, then $1/(1-\cos (j + \pi)) \approx 1/(1-(1 - \epsilon^2/2))= 2/\epsilon^2$. Heuristically speaking, we might expect this to happen every $1/\epsilon$ integers, so on average these spikes would not destroy the long-term decay with $\ln^{-k-1}(n_0)$ (which can be made quantitively, though not qualitatively, stronger by increasing $n_0$). However, I don't know how to make this argument rigorous, or whether that is even possible given the "unpredictability" of $\pi$. It's interesting that this proof might depend on the details of $\pi$ in this way. [Update: Thanks to George Lowther for the comments pointing to this MathWorld page, which gives an upper bound on the irrationality measure of $\pi$ (never heard of that before :-). That this is finite implies that $M_k$ can be bounded by a power of $k$, and hence the outer sum converges (uniformly) -- I believe that completes the proof.] 
Assuming that the spike problem can be solved, we still have the interchangeability of the sums to come back to. This theorem shows that we may interchange the sums if the convergence of
\[\sum_{k=0}^\infty \sum_{n=n_0}^m \frac{(-1)^n}{\ln(n)} \frac{\cos^k(n)}{\ln^k(n)}\]
is uniform in $m$ -- which it is, since $m$ dropped out in the derivation of the bound $M_ka_{n_0}$.
To summarize, this derivation shows that the series can be shown to converge if the irregular spikes in $1/(1-\cos (j + \pi))$ that occur because $j$ gets arbitrarily close to odd multiples of $\pi$ can be shown not to destroy the convergence of the outer sum over $k$.
Thanks for an interesting question :-).
A: Inserting \sum_{n=2}^{infty} (-1)^n/(ln(n)+cos(n)) into the search box of Wolfram Alpha gives an approximate value of –4.552162944044021. However unbelievable the actual sum may be, you can be a bit sure about its convergence.
