Convergence of $\sum_{k=1}^{\infty} \left (\sum_{j=1}^{k}\frac 1 j\right)^{-k}$ 
Does $\displaystyle\sum_{k=1}^{\infty} \left (\sum_{j=1}^{k}\frac 1 j\right)^{-k}$ converges ?

Let's call the inner sum $a_k$ such that $\displaystyle\sum_{k=1}^{\infty} (a_k)^{-k}$, applying root test we get: $(a_k)^{-1}= \left(\sum_{j=1}^{k}\frac 1 j\right)^{-1} = \frac {1}{1/1+1/2+...1/k}<1$ 
So the given sum converges. Is that all right ?
 A: Starting with the second term, the $k$th term is at most $(1 + 1/2)^{-k} = (2/3)^k$, so since the geometric series $\sum_{k=2}^{\infty} (2/3)^k$ converges so does the original series. 
A: A nice idea in case you encounter the harmonic series again:
notice that $\sum_{k=1}^{\infty} \left (\sum_{j=1}^{k}\frac 1 j\right)^{-k}$ is convergent, as is $ \sum_{k=2}^{\infty} \frac{1}{( \ln k)^k}$ because $\sum_{j=1}^{k}\frac 1 j \ge \ln k$ 
Now since $\ln k >1$ for $k>3$ the series obviously converge by comparison with the geometric series with general term $\frac{1}{ (\ln 4)^n}$.

To answer your comment:
In a sense, yes. First check that 
$$\frac{1}{k}=\int_{k}^{k+1}\sup_{x \in [k,k+1]}(\frac{1}{x})dt \ge \int_{k}^{k+1} \frac{dt}{t} \ge \int_{k}^{k+1}\inf_{x \in [k,k+1]}(\frac{1}{x})dt = \frac{1}{k+1}.$$ 
Then calculating the integral in the middle you get $$\int_{k}^{k+1} \frac{dt}{t}=\ln(k+1) - \ln(k) .$$
Summing things up to $N$ we obtain:
$$\sum_1^N \frac{1}{k} \ge \sum_{1}^N(\ln(k+1)-\ln(k))\ge \sum_1^N \frac{1}{k+1}.$$ To finish things, notice that you can telescope the expression in the middle and get $\ln(1+N) - \ln(1)=\ln(N+1)$.
