The sum of the series frac{1}{2!} - frac{1}{3 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) We know the Taylor series expansion for $e^x$ is given by:$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots$Substitute $x =

Vedclass · Accepted Answer

$e^{-1}$

Vedclass · Answer

(D) We know the Taylor series expansion for $e^x$ is given by:$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots$Substitute $x = -1$ into the expansion:$e^{-1} = 1 + (-1) + \frac{(-1)^2}{2!} + \frac{(-1)^3}{3!} + \frac{(-1)^4}{4!} + \dots$Simplifying the terms:$e^{-1} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$Since $1 - 1 = 0$,we get:$e^{-1} = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$Thus,the sum of the series is $e^{-1}$.

The sum of the series $\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$ up to infinity is:

Similar Questions

Sum the series to infinity $\frac{3}{4} - \frac{5}{4^2} + \frac{3}{4^3} - \frac{5}{4^4} + \frac{3}{4^5} - \frac{5}{4^6} + \dots$

If the $12^{th}$ term of an $A.P.$ is $-13$ and the sum of the first four terms is $24,$ then what is the sum of the first $10$ terms?

Find the sum of all odd numbers of four digits which are divisible by $9.$

The sum of the first $n$ terms of the series $1^2 + 2.2^2 + 3^2 + 2.4^2 + 5^2 + 2.6^2 + \dots$ is $\frac{n(n + 1)^2}{2}$ when $n$ is even. When $n$ is odd,the sum will be:

The sum of $24$ terms of the following series $\sqrt 2 + \sqrt 8 + \sqrt {18} + \sqrt {32} + \dots$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module