What is the sum from k=1 to infinity of cos(1/k)-cos(1/(k+1)) ?

The sum from k=1 to infinity of cos(1/k)-cos(1/(k+1)) is cos(1)-1

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sum from k=1 to infinity of cos(1/k)-cos(1/(k+1))

k = 1 \sum \infty cos (\frac{1}{k}) - cos (\frac{1}{k + 1})

cos (1) - 1

Solution steps

k = 1 \sum \infty cos (\frac{1}{k}) - cos (\frac{1}{k + 1})

Expand telescoping series:

cos (1) - 1

= cos (1) - 1

n = 0 \sum \infty \frac{( 3 ^{n} )}{2}

\int \frac{9}{x ^{2} ( x ^{2} + 25 )} d x

\frac{1}{x}

\frac{d x}{d t} = 6 - \frac{3 x}{500}

\frac{( 6 s - 4 )}{s ^{2} - 4 s + 20}

What is the sum from k=1 to infinity of cos(1/k)-cos(1/(k+1)) ?
The sum from k=1 to infinity of cos(1/k)-cos(1/(k+1)) is cos(1)-1