# Logarithmic series

Can you give the answer?

bro the T_r term has a sigma in the numerator and no sigma in the denominator. The T_r term then gets whole sigma(ed) to get the summation. So the sigma in the T_r term is \sum_{i=1}^n 2^n. This gets simplified to 2^{n+1}-1 (ofc divided by the factorial term). This gives us T_r which further gets summed up to yield the required summation.

Bro but why isnâ€™t sigma in Dr? It should be there right? Then only itâ€™ll be equal to the question.

One thing I noticed is in question itâ€™s till infinity, but you have taken till nâ€¦some logic Iâ€™m missing out?

(Sorry, if Iâ€™m asking a dumb question)

In the term we donâ€™t need to take sum the denominator bro coz observe that only in the numerator there is summation in every term. The denominator has a simple factorial. And I have considered it till infinity only. The e^x expansion holds true only when we consider n to be infinity.

Although I do realize that I havenâ€™t specifically mentioned infinity anywhere. Jus out of hurry coz Iâ€™ve got my english exam lol

Really dumb â€¦thanks bro!! You rockâ€¦

Just read the English exam partâ€¦All the best bro!! Gr88â€¦solving maths doubt in between eng prep