Find highest $m$ such that $\sum\limits_{n=0}^m 9(n+1)10^n\le10^i$ for very large $i$

Question

Consider a sum $$\sum_{n=0}^m 9(n+1)10^n$$ and a big power of 10, like $$p=10^{100000}$$ 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.

Answer 1

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 $m(k)$.

Answer 2

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