WBJEE · Maths · Limits
Let \(t_{n}\) denotes the nth term of the infinite series \(\frac{1}{1 !}+\frac{10}{2 !}+\frac{21}{3 !}+\frac{34}{4 !}+\frac{49}{5 !}+\ldots .\) Then \(\lim _{n \rightarrow \infty} t_{n}\) is
- A \(e\)
- B \(0\)
- C \(e^{2}\)
- D 1
Answer & Solution
Correct Answer
(B) \(0\)
Step-by-step Solution
Detailed explanation
Let \(\quad S=1+10+21+34+49+\ldots+t_{n}^{\prime}\) and \(S=1+10+21+34+\ldots+t_{n}^{\prime}\) \(0=1+9+11+13+15+\ldots-t_{n}^{\prime}\)…
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