Is there any possibility to do divergent summation with $\sum_{k=1}^{\infty}\exp(\sqrt k) $?

Solution 1:

I think you want Ramanujan summation. (and I believe it gives you the same answers you've already got)

Solution 2:

Hmm, I'm answering myself with the best hypothesis - I think I got it now.

+++I had to correct the computation-procedure, all earlier data are replaced+++

More insight was possible after generalization of the problem, where I do not only consider the square-root of the running exponent but just a fractional power $q$ : $$S(q) = \exp(1^q) + \exp(2^q) + \exp(3^q) + \ldots \ \ \ \ 0<=q<=1 \tag 1$$ Then we have also two parameters $q$ for which solutions are known: for $q=1$ we have the geometric series with quotient $e$ and for $q=0$ we have $e*\zeta(0)$
$$ S(0) = e^1 + e^1 + e^1 + e^1 + \ldots = e*\zeta(0) \approx -1.3591 \tag 2$$ $$ S(1) = e^1 + e^2 + e^3 + e^4 + \ldots = \frac{e}{1-e} \approx -1.5819 \tag 3$$ I'll denote $$f_q(x) = exp(x^q) = \sum_{k=0}^{\infty} \frac{x^{q*k}}{k!} \tag 4$$ and $$g_q(x)= \exp((1+x)^q-1)=\exp((1+x)^q)/e =f_q(x+1)/e = \sum_{k=0}^{\infty} g_k\frac{x^{q*k}}{k!} \tag 5$$ .

The Ramanujan-summation approximates that values well, so I use this as reference solution. For notational easiness I rewrite the wikipedia formula in terms of zeta instead of Bernoulli numbers and the derivatives of $f$ cancelled by the factorials such that just the coefficients $f_k$ remain; so $$ C(a) = \int_0^a f_q(t)dt + \sum_{k=0}^{\infty} \zeta(k)f_k \tag 6$$ Denote the integral with a name: $c_{q,a} =\int_0^a f_q(t)dt$

Now because the derivative of $f$ at zero is infinite we can use the representation by the g-coefficients: $$ C(a) = c_{q,a} + e*\sum_{k=0}^{\infty} \zeta(k)g_k \tag 7$$ Now we should use $C(0)$ for the Ramanujan-sum, so $c_{q,a}$ vanishes and $$ R(q) = e*\sum_{k=0}^{\infty} \zeta(k)g_k \tag 8$$

Unfortunately the sum does not yet converge, so that expression must again be done with a divergent summation-procedure. With that summation I get good approximations for the reference values $$ \begin{array} {rlll} R(0) &\approx& -1.3591...&=& S(0)\\ R(0.5) &\approx& -1.4719988...&???& S(0.5) \\ R(1) & \approx & -1.5819... &=& S(1) \end{array} \tag 9$$ where $R(0.5)$ is a new value for the divergent sum $ S(0.5)=\sum_{k=1}^{\infty} e^{\sqrt k}$.

--

The divergent sum of the $R()$-computation can be replaced by the adaption of my first described summation-method. By changing the summation-order in the double-series for $S(q)$ I arrive at a formal expression $$s_0(q) = \sum_{k=0}^{\infty} \frac{\zeta(-kq)}{k!} \tag {10}$$
With this Pari/GP gives me about $ s_0(1/2)=-0.753717005339 $ .

Two corrections seem to identify this with the $R(q)$-sums; the integral $c_a$ and one instance of $e$: $$ G(q) = c_{1,q} - e + s_0 \tag {11}$$

Unfortunately I've not yet understood, why that specific corrections are required.

-- Here is a plot for $s(q)$ where $q=0 \ldots 1.5 $ (the previously displayed plot had wrong data and is replaced)

(The line is not perfectly linear)