Harmonic Series question about convergence

Question

For large enough $n \in \mathbb{N}$, consider the sequence $(a_i (n))_{i \in \mathbb{N}} \overset{\Delta}{=} (a_i)_i : a_i(n) \overset{\Delta}{=} a_i = \frac{\sum_{x=1}^n (\frac{1}{x^i})}{i} \forall i \in \mathbb{N}$. Is (a_i)_i monotonic decreasing to $0$ (again, for any sufficient large $n$, that is)?

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Here is my real question: Does the following series converge: $-\frac{\sum_{x=1}^n (\frac{1}{x^1})}{1} + \frac{\sum_{x=1}^n (\frac{1}{x^2})}{2} - \frac{\sum_{x=1}^n (\frac{1}{x^3})}{3} + \frac{\sum_{x=1}^n (\frac{1}{x^4})}{4} - \frac{\sum_{x=1}^n (\frac{1}{x^5})}{5} + \dots$? If so, to what value? (It should be dependent on $n$, I believe).

Alternatively, why is $\underset{i \rightarrow \infty}{\lim} (\frac{\zeta(i)}{i}) = 0$ (and monotonically so)?

My thoughts: I see the geometric series in the coefficients of each term. Could I differentiate each sum $\frac{\sum_{x=1}^n (\frac{1}{x^i})}{i}$ by $i$ in order to play around with things?

Answer 1

Here is an answer to your question " why is $\lim_{i \to \infty} \zeta(i)/i = 0$ ( and monotonically so)."

The infinite series representation of $\zeta(s)$ converges uniformly for $\Re s \geq 1 + \delta.$

Consequently, we can interchange limits,

$$\lim_{s \to \infty} \zeta(s) =\lim_{s \to \infty} \lim_{m \to \infty}\sum_{n=1}^{m} n^{-s} = \lim_{m \to \infty} \lim_{s \to \infty}\sum_{n=1}^{m} n^{-s}= \lim_{m \to \infty} \lim_{s \to \infty} [1 + 2^{-s} + 3^{-s} + \ldots] = 1.$$

Hence,

$$\lim_{s \to \infty} \frac{\zeta(s)}{s} = 0.$$

If $s_2 > s_1$, then $n^{-s_2} < n^{-s_1}$, and

$$\sum_{n=1}^{m} n^{-s_2} < \sum_{n=1}^{m} n^{-s_1}.$$

Since this is true for all $m$, we have

$$\zeta(s_2) < \zeta(s_1),$$

and

$$\frac{\zeta(s_2)}{s_2} < \frac{\zeta(s_1)}{s_1}.$$

Therefore, both $\zeta(s)$ and $\zeta(s)/s$ are monotonically decreasing.

Proof of limit interchange.

For a partial sum, given any $\epsilon > 0$ there exists $K(m) > 0$ such that if $s > K(m)$ , then

$$\left|\sum_{n=1}^m n^{-s} - 1 \right| < \frac{\epsilon}{2}.$$

By uniform convergence, there exists $M \in \mathbb{N}$ such that if $m \geqslant M$ then for all $s$ with $\Re s \geq 1 + \delta,$

$$\left|\zeta(s) - \sum_{n=1}^m n^{-s}\right| < \frac{\epsilon}{2}.$$

Hence, if $s > K(M)$, then

$$|\zeta(s) - 1| \leqslant \left|\zeta(s) - \sum_{n=1}^M n^{-s}\right|+ \left|\sum_{n=1}^M n^{-s} - 1 \right| \leqslant \epsilon.$$

Answer 2

$$\sum_{j=1}^\infty \dfrac{(-1)^j x^{-j}}{j} = - \log(1 + 1/x) = \log(x) - \log(x+1)$$ converging for $x \ge 1$. Sum this for $x = 1$ to $n$ and you get $-\log(n+1)$.