sum of the series from n =1 to ∞ whose nth term is 1/(n+1)! a) e b) e-2 c)e+1 d) e-3

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## 1 Answer

We know the derivative of eˣ is eˣ.

If we write eˣ=a₀+a₁x+a₂x²+...+a_nx^n

When x=0, 1=a₀.

Now differentiate:

eˣ=a₁+2a₂x+...+na_nx^(n-1)

and put x=0:

1=a₁.

Differentiate again:

eˣ=2a₂+...n(n-1)x^(n-2).

When x=0, 1=2a₂.

If we repeatedly differentiate, and put x=0 to solve for a_n, we end up with the series:

1+x+x²/2!+x³/3!+...+x^n/n! to infinity.

Put x=1 and e=1+1+1/2!+1/3!+...+1/n!.

e-2=1/2!+1/3!+...

So the general term is 1/(n+1)!, the required series.

Therefore (b) is the answer.

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