# A “fast” way to find the sum of the sequence $5,5.5,5.55,5.555,5.5555,\ldots$ (20 terms)

My initial approach is diving the whole sum by $9$ and taking the common $5$ out which gives $$\frac{5}{9}[(10-1)+(10-0.1)+(10-0.01)+\cdots + (10-10^{-19})]$$ after some algebra this could be reduced to $$\frac{5}{9} \times [200-\frac{10}{9} \times (1-10^{-20})]$$

after this I am not sure how to show that is almost equal to $110.5$? Also if any body wants to suggest any other tricky/fast way I will appreciate it.

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Hard to find an improvement on this method, the last step being to type 8950/81 in your favorite calculator. – Did Aug 16 '11 at 10:19
@Didier Piau: Heck, you can do the division in your head easily enough (at least far enough to note that 8950.5/81 = 110.5 exactly). – Ilmari Karonen Aug 16 '11 at 13:52
You found a very efficient method. For the last step, note that $10^{-20}$ is utterly negligible in comparison with the other numbers in your expression. If you replace it by $0$, the distance from the truth is only $(50/81)\times 10^{-20}\:$! In fact, you can then write down the exact answer, since the last digit must be $5$. – André Nicolas Aug 16 '11 at 15:50
@André:Thanks,I guess your comment is more than enough for me :-) – VelvetThunder Aug 16 '11 at 16:02
@Ilmari: Good remark. I guess I was thinking about getting 5-6 decimals. – Did Aug 16 '11 at 21:04

$5+5.5+5.55+5.555+\cdots$

$= 5 + (5+0.5) + (5+0.5+0.05) + \cdots$

$= 20 \times 5 + 19 \times 0.5 + 18 \times 0.05 + 17 \times 0.005 \cdots + 1 \times 0.00{\ldots}05$

$\approx 100+9.5+0.9+0.085$

$= 110.485$.

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Would you please like to add few more lines as explanation? – VelvetThunder Aug 16 '11 at 16:10
@FoolForMath, done. – lhf Aug 16 '11 at 16:17
+1 and accepted.This seems to be faster that my approach and also purports to be a general way,thanks :-) – VelvetThunder Aug 16 '11 at 16:22

A slower way. It solves the general case with $n$ terms. I denote your sum as $$S_{20}=5+5.5+5.55+5.555+\ldots +5.\underset{19}{\underbrace{555\ldots 5}}$$

and the general case for $n$ as

$$S_{n}=5+5.5+5.55+5.555+\ldots +5.\underset{n-1}{\underbrace{555\ldots 5}}.$$

Since $\sum_{j=1}^{k-1}10^{j-1}=\frac{1}{90}10^{k}-\frac{1}{9}$, we have in general

$$\begin{eqnarray*} S_{n} &=&5n+5\sum_{k=1}^{n}\frac{\sum_{j=1}^{k-1}10^{j-1}}{10^{k-1}} =5n+5\sum_{k=1}^{n}\frac{\frac{1}{90}10^{k}-\frac{1}{9}}{10^{k-1}} =\frac{1000}{9}-\frac{50}{81}+\frac{50}{81}10^{-n}, \end{eqnarray*}$$

and in the present case

$$\begin{eqnarray*} S_{20} &=&\frac{1000}{9} -\frac{50}{81}+\frac{50}{81}10^{-20} \\ &=&\frac{220987654320987654321}{2000\,000\,000000\,000\,000}\approx 110.49. \end{eqnarray*}$$

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Ehhh.. the last division is scaring me!I am supposed to solve this without calculator :-) – VelvetThunder Aug 16 '11 at 16:03
@FoolForMath: you only have to divide by 2. Take e. g. the number 220987654/2=110493827 and multiply by $10^{-6}$ – Américo Tavares Aug 16 '11 at 16:11
+1,Aha!I didn't notice that before. – VelvetThunder Aug 16 '11 at 16:15
@FoolForMath: Do you mean $(50/81)10^{-20}$? – Américo Tavares Aug 16 '11 at 16:15
I meant this:$\frac{220987654320987654321}{2000\,000\,000000\,000\,000}\approx 110.49$,but as I understand I don't need to compute up-to this extent/accuracy. – VelvetThunder Aug 16 '11 at 16:24

As an alternative you could try \begin{align} \sum_{k=0}^{19}(\frac{50}{9}-\frac{5}{9}10^{-k})&=\frac{50}{9}\times 20-\frac{5}{9}\frac{1-10^{-20}}{1-10^{-1}}\\ &=\frac{1000}{9}-\frac{50}{81}\left(1-10^{-20}\right)\\ &=\frac{8950}{81}+\frac{50}{81}\times 10^{-20}\\ &=110\frac{40}{81}+\frac{50}{81}\times 10^{-20} \end{align}

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Sorry but I fail to see how this is an alternative to @Fool's method. This is exactly Fool's method. – Did Aug 16 '11 at 10:55
@Didier Piau: I guess I didn't notice where FoolForMath showed a method other than "after some algebra...". Since he was looking for something faster, I figured that whatever he did was not as fast. He also wanted to see why this was almost $110.5$, so I showed it was close to $110\frac{40}{81}$. If FoolForMath says this adds no value, I will delete my answer. – robjohn Aug 16 '11 at 11:11