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

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!

• If $n>0$, then $n^{1/2^j}\rightarrow 1$ as $j\rightarrow\infty$, so the series does not converge as $k\rightarrow\infty$. – J.R. Feb 5 '14 at 10:16
• Converge in what sense? The terms approach 1 as $k\rightarrow \infty$ – gammatester Feb 5 '14 at 10:18
• Okay, edited the question.. I actually just want to evaluate the Sum of this series. – sanjeev mk Feb 5 '14 at 10:19
• So the sum of the terms does not converge, indeed. – Ian Coley Feb 5 '14 at 10:19
• Not sure if this helps entirely, since it involves the limit of your sum, but there is some kind of convergence here: wolframalpha.com/input/… (WolframAlpha couldn't calculate it for a general $n$) – Ian Coley Feb 5 '14 at 10:30

[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]

• $s_n(a,b)$ is the $n$-th coefficient of the series? – Martín-Blas Pérez Pinilla Feb 5 '14 at 15:49
• @Martín: no, the sum of the finite series. In the OP's question it is also $a=0$ to have things simpler. – Gottfried Helms Feb 5 '14 at 15:52
• Then, why there is a power series? You are saying that some (not very well known) function is analytic. – Martín-Blas Pérez Pinilla Feb 5 '14 at 15:53
• The power series is $h(x)$. You see that this power series is involved with the parameters of the powers of n as given in the OP's question. It might remind of the Bernoulli-polynomials/Faulhaber-formula with which you also sum finite sums (of like-powers of course) in a very similar way, only that the Bernoulli-Polynomials/Faulhaber-formula is a finite polynomial and here the function $h(x)$ is a power series, having an infinite number of terms. – Gottfried Helms Feb 5 '14 at 15:55

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.

• He's not asking if it converges. He's asking about finite sums. – Ian Coley Feb 5 '14 at 10:34

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 $n$ $$n-n^{1/2^{k+1}}\leq R(n,k)-(k+1)\leq n-1.$$