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

Question

Self-studying some properties of the exponential-function I came to the question of ways to assign a value to the divergent sum $$s=\sum_{k=1}^{\infty}\exp(\sqrt k) $$ I have no idea how to attack this with standard methods (I do not know many).

I tried a replacement using reversing order of summation when $s$ is expressed as double sum and introducing zeta at negative half-integer values. I wrote the double-sum as $$ \begin{array} {rr} s &=& \sum_{k=1}^{\infty} \sum_{j=0}^{\infty} \frac{k^{j/2}}{j!} &=& \sum_{j=0}^{\infty} \frac{\sum_{k=1}^{\infty} k^{j/2}}{j!} &=& \sum_{j=0}^{\infty} \frac{\zeta(-j/2)}{j!} \end{array}$$

With this Pari/GP gives me about $ s=-0.753717005339 $ .

Moreover I'm curious whether I have to extend the index to $-2$ to get formally $$s = -1 + 0 + \sum_{k=0}^{\infty}\frac{\zeta(-k/2)}{k!}$$ where I've set $\frac{\zeta(0.5)}{(-1)!} = 0 $ and $\frac{\zeta(1)}{(-2)!} =\frac{(-1)!}{(-2)!} =-1 $ to arrive at $ s_1=-1.753717005339 $

But this is just a shot in the dark. I've tried with another trick which works in some circumstances to sum the alternating series first, and then add an alternating partial series etc, like $$s=\sum_{k=1}^{\infty} (-1)^{k-1} \exp(\sqrt k) + 2*\sum_{k=1}^{\infty} (-1)^{k-1} \exp(\sqrt{ 2 k}) + 4*... $$ but this doesn't help either since all sums are positive and I get another divergent and monotonuously increasing sequence of partial sums. One of the problems is, that regular summations cannot sum divergent series if all summands are positive and also increase, so my standard tools fail here.

Q: how else could I sum that series?

[edit]: Things begin to complicate... I tried another approach and got a result with a suspicious integer difference. I look at the formal powerseries $$ g(x) = \exp(\sqrt{1+x}-1) = 1 + g_1 \frac{x}{1!}+ g_2 \frac{x^2}{2!}+ \ldots $$ and $$ \begin{array} {ll} t&=&e*(g(0)+g(1)+g(2)+g(3)+...) \\\ &=& e*g(0) + e*(g(1)+g(2)+g(3)+\ldots) \\\ &=& e*g(0)+e*t_1 \end{array} $$
Then, by the same principle of reordering summation of the formal doubleseries we get another sum of zetas, but now at negative integer arguments $$ \begin{array} {ll} t_1 &=& g(1)+g(2)+\ldots \\\ &=& 1*\zeta(0)+g_1*\frac{\zeta(-1)}{1!} +g_2*\frac{\zeta(-2)}{2!}+\ldots \end{array} $$ if that reordering makes sense.
Interestingly the value which I get by this is $t=1.246282994682$ where much interestingly $s=t-2$ . This does not yet confirm one value over the other. But to have just a simple integer-difference seems to tell, that in principle these paths of computation are not completely meaningless (?)

[Edit2]: using Ramanujan-summation as shown in the wikipedia-link I arrive at the same latter value of about $1.2462$ which is again $s+2$. What puzzles me is, that I'm used to negative values for divergent sums of increasing positiv terms. Did I miss something in the Ramanujan-formula? I used the formula $$C(a) = \int_0^a f(t) dt - \frac12 f(0) - \ldots $$ where I insert my $g(x)$ above for $f(x)$ using $a=0$ (and thus the integral-term being zero). [Edit3]: It seems, that the method at [edit1] is just the Ramanujan-method where the Bernoulli-numbers are translated to the respective zeta-values [Edit4]: I had to correct the sum-formula; the factorials had to be removed
Then the wikipedia-formula reads $$C(a) = \int_0^a f(t) dt + \sum_{k=0}^{\infty} \zeta(-k) \frac{f^{(k)}(0)}{k!} $$ Because we have a power series, the k'th derivative $f^{(k)}(0)=f_k*k! $ and we can replace this in the formula, cancelling the factorial. Furtherly I had used the function g(x), so $$ t=e + e*\sum_{k=0}^{\infty} \zeta(-k) g_k $$ should equal $s$ . Unfortunately, this is also a divergent sum, but can be Euler-/Borel-summed.
Interestingly, the integral-term $ \int_a^b f(t) dt $ gives just the mysterious number 2 : $$ e*\int_{-1}^0 g(t) dt = \int_{0}^1 \exp(\sqrt{t}) dt = 2 $$ according to wolfram-alpha .
With this it seems I can use the much better summable (if not convergent) series $$ s=\sum_{j=0}^{\infty} \frac{\zeta(-j/2)}{j!} $$ and determine that Ramanujan-summ $$ t = \int_{0}^1 \exp(\sqrt{t}) dt + s $$

[Edit5] It seems, that the possibility for computations is coherent for other exponents in the basic series. If I generalize $$ \begin{array} {rr} S_q &=& \sum_{k=1}^{\infty} \exp(k^q) \\ g_q(x) &=& \exp((1+x)^q-1) &=& \exp((1+x)^q)/e \\ Ca_q &=& \int_0^1 \exp(x^q) \end{array} $$ and $t_q$ and $s_q$ accordingly, then for some positive fractional q I get the following results. $$ \small{ \begin{array} {rrrr} q& t_q & Ca_q & & s_q & C_q(0) \\ 1.00000000000 & 1.13630512159 & 1.71828182846 &(= 1e - 1) & -0.581976706869 & 1.13630512286 \\ 0.500000000000 & 1.24628299466 & 2.00000000000 &(=- 0e + 2)& -0.753717005339 & 1.24628299491 \\ 0.333333333333 & 1.28422772983 & 2.15484548538 &(= 3e - 6)& -0.870617755549 & 1.28422772997 \\ 0.250000000000 & 1.30316006154 & 2.25374537233 &(=-8e + 4!) & -0.950585310784 & 1.30316006162 \\ 0.200000000000 & 1.31447347236 & 2.32268228066 &(=45e - 5!)& -1.00820880830 & 1.31447347240 \\ 0.166666666667 & 1.32198952281 & 2.37359728681 &(=-264e + 6!)& -1.05160776400 & 1.32198952284 \\ 0.142857142857 & 1.32734318867 & 2.41279179153&(=1855e - 7!) & -1.08544860285 & 1.32734318869 \\ 0.125000000000 & 1.33134949700 & 2.44392029544 &(=-14832e + 8!)& -1.11257079844 & 1.33134949702 \\ 0.111111111111 & 1.33445988876 & 2.46925379716 &(=1334978e - 9!)& -1.13479390840 & 1.33445988878 \\ 0.100000000000 & 1.33694450266 & 2.49028031297 &(=-A240(10)e + 10!)& -1.15333581032 & 1.33694450267 \\ 0.0909090909091 & 1.33897484256 & 2.50801667035 &(=A240(11)e - 11!)& -1.16904182779 & 1.33897484257 \\ 0.0833333333333 & 1.34066501173 & 2.52318189730 &(=-A240(12)e + 12!)& -1.18251688557 & 1.34066501173 \end{array} } $$ where $t_q$ in the second column is the assumed sum computed by the method $ t_q = Ca_q + s_q $ . The $A240(k)$-entries are also found in the sequence A000240 in OEIS beginning at $k=1$.

The last column is the same result computed by the Ramanujan sum $C(0)$ as denoted in the wikipedia-article (and my translation into the $g_q()$-function). That terms are always a diverging sequence of partial sums, so their Euler-sum is documented here for comparision of accuracy.

A plot of $q$ and $t_q$ looks like a nearly linear (negative) relation.

It is still open, which value ($t_q$ or $s_q$) should be taken as final sum.

Note that using $q=1$ we should get $ S_1 = e^1 + e^2 + e^3 + ... = \frac{e}{1-e} \approx -1.58197670687 $ where only $s_q$ is in the near (misses by 1).
The $t_q$ value for $q=1$ however seems obscure; the correct value would be $ S_1 = t_q - e = \frac{e}{1-e} $ . Here I do not know what this tells us?

Answer 1

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

Answer 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)