Summation evaluation of $\sum_{j=0}^k n^{1/2^j}$?

Question

How do I go about solving this:

$\sum_{j=0}^k n^{1/2^j}$

So, the terms of this series are $n , n^{1/2},n^{1/4},n^{1/8},.......n^{1/2^k}$

Any insights on what the thought process should be, to solve this?

Any links to resources or even the name of this series would also help..

I initially thought the Power Series or the Puiseux Series look similar to this, but I don't think they apply to this..

Clarification:

I'm trying to find a relation $R(n,k)$ for the finite sum of this series.

Any help is appreciated, thanks!

Answer 1

[update] There is a power-series which allows to compute $$ s_n(a,b) = n^{1/2^a} + n^{1/2^{a+1}} + ... + n^{1/2^b} \tag 1$$ (even with fractional bounds of the interval) .

This can be dealt with much related to the idea behind the Bernoulli/Faulhaber-polynomials for summing of consecutive like-powers - only that we do not have a (finite) polynomial but a power- series (with infinitely many terms)

Don't know, whether this is helpful and in the direction where you want to go to, just a viable way to go...

By inspection of all matrices with which I'd computed my first version (for reference I've left the text at the end of my answer) I've arrived at the following solution

We can write another expression for the sum $s_n(a,b)$ in the following form: $$ s_n(a,b) = \sum_{k=1}^\infty \left( {\log(n)^k \over k!} \cdot { A^k-B^k \over 1-Q^k } \right) + (b+1-a) \\ \\ \text{ where } Q={1 \over 2}, A=Q^a , B=Q^{b+1} \tag 2 $$

Implemented in Pari/GP this is

\\ Pari/GP-code
{mysum(n,a,b) = local(Q,A,B,L,s); 
    Q=1/2; A=Q^a;B=Q^(b+1);L = log(n);
    s=sumalt(k=1,L^k/k!* (A^k-B^k)/(1-Q^k));   \\ "sumalt" is a summation
                                 \\ procedure in Pari/GP which mimics summation
                                 \\ of infinite series and even can sum many not
                                 \\ too strong diverging alternating series as it
                                 \\ is the case here
    s=s + ((b+1) - a);
    return(s); }


\\ Test with some parameters
[a=2,b=5,n=2]
sum(k=a, b, n^(1/2^k) )  \\ the sum as stated in the OP's question   
 %1436 = 4.345885779

mysum(n,a,b)             \\ the sum using the power-series
 %1437 = 4.345885779


[a=5,b=17,n=3]

sum(k=a, b, n^(1/2^k) )  \\ the sum as stated in the OP's question   
 %1439 = 13.06944843

mysum(n,a,b)             \\ the sum using the Puisieux-series
 %1440 = 13.06944843

[/update]

---

[my initial text to which Martín's comments are related]

There is a power series (with a not-yet determined range of convergence) which allows to compute $$ s_n(a,b) = n^{1/2^a} + n^{1/2^{a+1}} + ... + n^{1/2^b} \tag 1$$

Unfortunately a simple formula for the coefficients are not yet known to me, I just computed them with the help of Carlemanmatrices.

The empirically found powerseries defining a function $h(x)$ begins as $$h(x) = 2 x - 1/3 x^2 + 0.19047619... x^3 - 0.13015873... x^4 \\ + 0.097491039... x^5 - 0.077237299... x^6 + 0.063555496... x^7 + O(x^8) \\ \tag 2 $$

and allows to formulate

$$ \begin{eqnarray} s_n(a,b) &=& n^{1/2^a} + n^{1/2^{a+1}} + ... + n^{1/2^b} \\ &=& h( n^{1/2^a} -1)- h( n^{1/2^{b+1}} -1) +(b+1-a) \end{eqnarray}\tag 3$$

The discussion and derivations in terms of the Carleman-matrices are a bit involved; we use also the concept of Neuman-series (for matrices) and I could present the path which is a rather general approach to many similar problems. However, after that much fiddling and reworking I've found that this can be simplified very much, see top of post...

[/initial text]

Answer 2

This is a clarification of commentaries.

If $\sum a_n$ converges, then $\lim a_n=0$. So if $\lim a_n=0$ is false (because does not exists or is $\ne0$) then $\sum a_n$ does not converges.

Answer 3

Let us write $n=\mathrm e^\alpha$, (nothing restricts $n$ to integers, actually). Let us evaluate $$R(n,k)-(k+1)=-(k+1)+\sum_{j=0}^kn^{2^{-j}}=\sum_{j=0}^k\left(\mathrm{e}^{2^{-j}\alpha}-1\right).$$ Let us expand it $$R(n,k)-(k+1)=\sum_{j=0}^k\sum_{m=1}^\infty\frac{\alpha^m}{m!}2^{-mj}=\sum_{m=1}^\infty\frac{\alpha^m}{m!}\frac{2^m-2^{-km}}{2^m-1}.$$ We can use here, for instance, $$ 1-2^{-m(k+1)}\leq \frac{2^m-2^{-km}}{2^m-1} \leq 1$$ and summing up we find $$ \mathrm{e}^\alpha-\mathrm{e}^{\alpha/2^{k+1}} \leq R(n,k)-(k+1)\leq \mathrm{e}^\alpha-1$$ or to go back to $n$ $$ n-n^{1/2^{k+1}}\leq R(n,k)-(k+1)\leq n-1.$$