Is there any possibilities that the following partial sum of the Dirichlet eta function can be zero?

Question

If we accept-- $s_o$ -- as one of the non-trivial zeros of the Riemann zeta function by $0 <Re(s_o)<1$ and $Re(s_o)$ is the real part of a complex variable, we know:

$$\eta(s_o ) = \sum_{1}^\infty \frac {(-1)^{n-1}} {n^{s_o}}=0 \tag{1}$$

Now, can we show that the following partial sum can not be zero?

$N$: Sufficiently large integer

$$\left|\sum_{1}^N \frac {(-1)^{n-1}} {n^{s_o}}\right|≠0 \tag{2}$$

Or can we show that the remainder term can not be zero?

$$\left|\sum_{N+1}^\infty \frac {(-1)^{n-1}} {n^{s_o}}\right|≠0 \tag{3}$$

Answer 1

$$\begin{eqnarray} f_s(t) = t^{-s}, f_s'(t) &=&-s t^{-s-1}, f_s''(t) = s(s+1)t^{-s-2} \\ f_s(n)-f_s(n+1) &=& -f_s'(n)+O(f_s''(n)) = s n^{-s-1} +O(s(s+1) n^{-s-2})\\ \sum_{n=N}^\infty (2n)^{-s}-(2n+1)^{-s} &=& \sum_{n=N}^\infty s (2n)^{-s-1} +O(s(s+1) (2n)^{-s-2}) \\ &=&s(s+1) 2^{-s-1}\sum_{n=N}^\infty \int_N^\infty x^{-s-2}dx+O(\frac{s(s+1)}{\Re(s+1)} (2N)^{-s-1}) \\ &=& s(s+1) 2^{-s-1} \int_N^\infty \lfloor x-N \rfloor x^{-s-2}dx + O(\frac{s(s+1)}{\Re(s+1)} (2N)^{-s-1}) \\ & =& s(s+1) 2^{-s-1} \int_N^\infty ( x-N +O(1)) x^{-s-2}dx + O(\frac{s(s+1)}{\Re(s+1)} (2N)^{-s-1}) \\ &=& s(s+1) 2^{-s-1} (\frac{N^{-s}}{s} - N\frac{N^{-s-1}}{s+1}+O(\frac{N^{-s-1}}{\Re(s+1)})) +O(\frac{s(s+1)}{\Re(s+1)} (2N)^{-s-1})\end{eqnarray}$$

which is non-zero for $N \ge N(s)$ large enough.

Answer 2

Can we show it as below for sufficiently large $N$ numbers? (We mean here if we speak about "the neighborhood of infinity" here; it is a subset of the set of real numbers which contains an interval ($M,+∞)$. It means this subset contains all sufficiently large $N$ numbers here as $N>M$).

Now let write the remainder term by accepting $´N´$ as even number here:

$$R_{N}(s) =\frac {1} {(N+1)^{s}}-\frac {1} {(N+2)^{s}} + \frac {1} {(N+3)^{s}}- \frac {1} {(N+4)^{s}}+⋅⋅⋅$$

Now, let us think two new functions as completely independent from the above-reminder term (it means they have no any connection with $R_{N}(s)$ )

$$P_{1}(s) =\frac {1} {(N)^{s}}-\frac {1} {(N+1)^{s}}+\frac {1} {(N+2)^{s}} - \frac {1} {(N+3)^{s}}+⋅⋅⋅$$

$$P_{2}(s) =\frac {1} {(N+2)^{s}}-\frac {1} {(N+3)^{s}}+\frac {1} {(N+4)^{s}} - \frac {1} {(N+5)^{s}}+⋅⋅⋅$$

If we handle them with $R_{N}(s)$ one by one:

$$R_{N}(s)+ P_{1}(s)=\frac {1} {N^{s}} \,\,\,\,\,\,\,\ and \,\,\,\,\,\,\,\ R_{N}(s)+ P_{2}(s)=\frac {1} {(N+1)^{s}} $$

Then let us consider a $´c´$ value as $c∈ℝ$ and $c≥1$. Thus, we can say that there exists a $´c´$ value, thus we can write the following conditions with all sufficiently large ´$N´$ numbers :

$$ |\frac {1} {c(N+1)^{s}}|<|2R_{N}(s)|<|\frac {c} {N^{s}}|$$

Thus, we conclude with sufficiently large ´N´ numbers at the neighborhood of infinity:

$$|R_N(s)| ≠0$$

But of course, we know $\lim_{N\rightarrow\infty} R_{N}(s)=0$.