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?
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: $$(1+N)\left(\log 5 + \frac{N}{2}\log 11\right)$$