2
$\begingroup$

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)?


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?

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.