# Is it possible to bound this sum?

I've got a sum:

$$\sum_{n=0}^m 9(n+1)10^n$$ And i have big power of 10, like $p=10^{100000}$ And i want to know what is the highest $m$ that this sum will be not greater than $p$? In fact i'm also interested in other powers of $10$. Is there some tool to help me with this? I know i can use computer for low powers of ten, but with such huge power it's not really possible.

I noticed that this sum is similiar to generating function for sequence $a_n = \langle 9 (n+1) \rangle$ and is equal to $$9\sum_{n=0}^\infty (n+1)x^n = \frac{9}{(1-x)^2}$$ I also tried to take $log_{10}$ on both sides, but i just can't make it work.

I'd appreciate some help on this! Thanks in advance

-

There are various ways to show that, for every $x\ne1$, $$\sum_{n=0}^m(n+1)x^n=\frac{(m+1)x^{m+2}-(m+2)x^{m+1}+1}{(x-1)^2},$$ hence, using this for $x=10$, one sees that the sums you are interested in are $$S_m=9\sum_{n=0}^m(n+1){10}^n=(m+\tfrac89){10}^{m+1}+\tfrac19.$$ To solve $$S_m=10^{10^k},$$ this suggests that one should look for $m$ such that $$(m+\tfrac89){10}^{m+1}\approx10^{10^k},$$ that is, to choose $m\approx m(k)$, where $$m(k)=10^k-k-1.$$ Finally, one can check directly that $$S_{m(k)}\lt10^{10^k}\lt S_{m(k)+1},$$ hence the largest value of $m$ such that $S_m\leqslant10^{10^k}$ is $m(k)$.

-
I think it's nice solution, but i have question, why $m(k)$ is equal to $10^k-k-1$? – Spinach Mar 30 '14 at 13:26
?? Because we define it as such. And I have a question myself: how is the accepted answer answering your query "to know what is the highest m that this sum will be not greater than p"? – Did Mar 30 '14 at 13:28
From first answer i got equation $$9S = 9x10^x - 10^x-1$$ for $x=m+1$ and i just automated it to have real answer. And i know that your answer is more precise and easier to calculate on paper, but i just didn't understand it ( sorry >:( ) fully – Spinach Mar 30 '14 at 13:39
"Automated it to have real answer" Meaning? But why accept not "real" answers? – Did Mar 30 '14 at 13:42
I meant i used computer to calculate this $m$, i know that the $S_m$ form are the same, but i didn't understand your way of coming to this form. And after that i didn't understand why $m(k) = 10^k - k - 1$ (why not $m(k) = 2*10^k - k - 1$ for example?) I meant i just didn't understand how to come to this, but i will surely change "accepted" mark to your post only if i could understand this bounding :-) Thanks for your posting @Did – Spinach Mar 30 '14 at 13:51

Let $\displaystyle S=9\sum_{n=0}^m(n+1)10^n=9\left[1+2\cdot10+3\cdot10^2+m\cdot10^{m-1}+(m+1)10^m\right]$

$\displaystyle\implies 10S=9\left[10+2\cdot10^2+3\cdot10^3+m\cdot10^{m}+(m+1)10^{m+1}\right]$

So, $\displaystyle10S-S=9(m+1)10^{m+1}-9\left[1+(2-1)10+(3-2)10^2+\cdots\{m+1-m\}10^m\right]$

$\displaystyle\implies9S=9(m+1)10^{m+1}-9\left(\underbrace{1+10+10^2+\cdots+10^m}\right)$

The terms under the brace is a finite Geomtric Series

Reference : Arithmetico-geometric sequence

-
Hey, thanks for your answer, but i don't really know how to continue from there (i just don't see it) – Spinach Mar 30 '14 at 12:42
@Chris, please find the edited version – lab bhattacharjee Mar 30 '14 at 12:45
I think it works great now... I think i understood the trick. Thanks:-) – Spinach Mar 30 '14 at 13:25
@Chris, My pleasure. Nice to hear that last statement – lab bhattacharjee Mar 30 '14 at 13:37