What is the sum from k=2 to infinity of 1/k-1/(4^k) ?

The sum from k=2 to infinity of 1/k-1/(4^k) is diverges

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sum from k=2 to infinity of 1/k-1/(4^k)

k = 2 \sum \infty \frac{1}{k} - \frac{1}{4 ^{k}}

diverges

Solution steps

k = 2 \sum \infty \frac{1}{k} - \frac{1}{4 ^{k}}

Apply Series Integral Test:

diverges

= diverges

(1 + ln (x) + \frac{y}{x}) d x - (1 - ln (x)) d y = 0

\frac{d y}{d x} = 2 x + y

y^{^{'}} + 2 x y = 0, y (0) = 3

\int sec^{6} (x) tan^{6} (x) d x

ln (x^{4} + 3)

What is the sum from k=2 to infinity of 1/k-1/(4^k) ?
The sum from k=2 to infinity of 1/k-1/(4^k) is diverges