 # Logarithmic series

1 Like

1 Like
1 Like
1 Like

Thanks a lot!!!
Bro but I didn’t understand 1->2 step…rest all I understood… 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.

1 Like

Bro but why isn’t sigma in Dr? It should be there right? Then only it’ll be equal to the question.

One thing I noticed is in question it’s till infinity, but you have taken till n…some logic I’m missing out?
(Sorry, if I’m asking a dumb question) 1 Like

In the term we don’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 haven’t specifically mentioned infinity anywhere. Jus out of hurry coz I’ve got my english exam lol

1 Like

Really dumb  …thanks bro!! You rock…  2 Likes

Just read the English exam part…All the best bro!! Gr88…solving maths doubt in between eng prep  1 Like

@Seshank I’m starting english tomorrow 2 Likes