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I have another question about series. Now, this is about series involving logarithm.

In the previous post, we can easily grouping the same factor from this series: http://tinyurl.com/kbg26ye

But, how about this? I think we need to use logarithm character to find the general formula, such that:

$$\log a +\log b= \log(a\cdot b)$$

Then:

$$\log 5+\log 5+\log 605+\log 6655+\dots = \log(5\cdot5) + \log(605\cdot6655)+....$$

But, it seems impossible to find the general formula from this problem. Because I cannot find the ratio for this problem. For example, the second term to the first term is differ by $1$, the third to the second is differ by $121$, but the last to the third is differ by $11$. It seems that there is no ratio for this problem.

Can anybody try to provide any idea to solve this problem?

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    $\begingroup$ What is the pattern of the arguments $5, 5, 605, 6655, \ldots$ you have in mind? $\endgroup$ – Travis Nov 30 '14 at 13:52
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    $\begingroup$ needs more terms for decipherance. $\endgroup$ – RE60K Nov 30 '14 at 14:06
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Assuming you meant the sum: $$\log 5 + \log 55 + \log 605 + \ldots = \sum_{n=0}^N \log\left(5\cdot 11^n\right)$$ Then this is equals to: $$\log \left(5^{N+1}\cdot 11^{0+1+2+\ldots+ N}\right)=\log \left(5^{N+1}\cdot \sqrt{11}^{N^2+N}\right)$$ Or after simplifying: $$(1+N)\left(\log 5 + \frac{N}{2}\log 11\right)$$

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    $\begingroup$ Good catch, indeed ! $\endgroup$ – Claude Leibovici Nov 30 '14 at 14:44
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    $\begingroup$ Why do complicated? $\sum_{n=0}^N \log(5\cdot 11^n) = \sum_{n=0}^N\bigl(\log(5)+n\log(11)\bigr) =\log(5)(N+1)+\log(11)\frac{N(N+1)}2$. $\endgroup$ – Marc van Leeuwen Nov 30 '14 at 15:03

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