JEE Mains · Maths · STD 12 - 7.2 definite integral
The value of \(\sum_{k=1}^{\infty}(-1)^{k+1}(\frac{k(k+1)}{k!})\) is :
- A \(2/e\)
- B \(1/e\)
- C \(\sqrt{e}\)
- D \(e/2\)
Answer & Solution
Correct Answer
(B) \(1/e\)
Step-by-step Solution
Detailed explanation
\(T_{k}=(-1)^{k+1}.\frac{k(k+1)}{\lfloor {k}}=(-1)^{k+1}(\frac{k(k-1)+2k}{\lfloor {k}})\) \(\therefore sum=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{\lfloor {k-2}}+\sum_{k=1}^{\infty}\frac{2(-1)^{k+1}}{\lfloor {k-1}}\)…
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