What is the sum from n=3 to infinity of (4^n+1)/(n!) ?

The sum from n=3 to infinity of (4^n+1)/(n!) is converges

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sum from n=3 to infinity of (4^n+1)/(n!)

n = 3 \sum \infty \frac{4 ^{n} + 1}{n !}

converges

Solution steps

n = 3 \sum \infty \frac{4 ^{n} + 1}{n !}

Expand

\frac{4 ^{n} + 1}{n !} : \frac{4 ^{n}}{n !} + \frac{1}{n !}

= n = 3 \sum \infty \frac{4 ^{n}}{n !} + \frac{1}{n !}

Apply the Sum Rule:

\sum a_{n} + b_{n} = \sum a_{n} + \sum b_{n}

= n = 3 \sum \infty \frac{4 ^{n}}{n !} + n = 3 \sum \infty \frac{1}{n !}

n = 3 \sum \infty \frac{4 ^{n}}{n !} =

converges

n = 3 \sum \infty \frac{1}{n !} =

converges

= converges + converges

= converges

y = \frac{1}{( 1 + x ^{2} )}

3 x^{- 2} - 4 x

e^{1 + (l n (x))^{2}}

1 - y^{2} d x - 1 - x^{2} d y = 0, y (0) = \frac{1}{2}

y^{^{'}} = e^{t - y}

What is the sum from n=3 to infinity of (4^n+1)/(n!) ?
The sum from n=3 to infinity of (4^n+1)/(n!) is converges