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

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

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sum from n=1 to infinity of (4n+1)/(4^{n+1)}

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

converges

Solution steps

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

Expand

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

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

Apply the Sum Rule:

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

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

n = 1 \sum \infty \frac{n}{4 ^{n}} =

converges

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

converges

= converges + converges

= converges

\frac{d}{d x} (e^{(- x^{2})})

\int sin^{3} (\frac{7}{3}) x d x

\int \frac{5}{9 - x ^{2}} d x

y^{^{''}} + 25 y = cos (5 x) + sin (5 x)

n = 1 \sum \infty \frac{7}{3 ^{n - 1}}

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