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Value of $\sum\limits_n x^n$

I have been working on this problem for awhile and have made efforts to try to resolve it. The question is: $$\sum_{n=1}^\infty 7(0.1)^{n−1}$$

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marked as duplicate by Marvis, EuYu, TMM, Old John, userNaN Nov 4 '12 at 22:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
thank you for responding, but i was asking if anyone knows how to find the partial sum of this expression. –  Nickeisha Cole Nov 4 '12 at 20:19
    
I don't see any question, but a formula! –  Mercy Nov 4 '12 at 20:21

1 Answer 1

For the "partial sum", I assume you meant $\sum_{n=1}^k 7(0.1)^{n−1}$ for any positive integer $k$.

Taking the $7$ out of the sum, and shifting the exponent, we get $7\sum_{n=0}^{k-1} 0.1^n$.

All this needs now is the standard formula $\sum_{n=0}^{k-1} x^n = (x^k-1)/(x-1) = (1-x^k)/(1-x)$. Putting $x = 0.1$, $\sum_{n=0}^{k-1} 0.1^n = (1-0.1^k)/0.9$.

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