TS EAMCET · Maths · Limits
\(\sum_{n=1}^{\infty} \frac{2 n}{(2 n+1) !}\) is equal to
- A \(\frac{1}{e}\)
- B \(\frac{e}{2}\)
- C \(e\)
- D \(2 e\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{e}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \sum_{n=1}^{\infty} \frac{2 n}{(2 n+1) !}=\sum_{n=1}^{\infty} \frac{2 n+1-1}{(2 n+1) !} \\ &=\sum_{n=1}^{\infty}\left(\frac{1}{2 n !}-\frac{1}{(2 n+1) !}\right) \\ &=\left(\frac{1}{2 !}-\frac{1}{3 !}\right)+\left(\frac{1}{4 !}-\frac{1}{5 !}\right)+\ldots \\…
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