What is the limit as n approaches infinity of \sum_{i=1}^n 1/(2n+i) ?

The limit as n approaches infinity of \sum_{i=1}^n 1/(2n+i) is ln(3)-ln(2)

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limit as n approaches infinity of \sum_{i=1}^n 1/(2n+i)

n \to \infty lim (i = 1 \sum n \frac{1}{2 n + i})

ln (3) - ln (2)

Decimal

0.40546 \dots

Solution steps

n \to \infty lim (i = 1 \sum n \frac{1}{2 n + i})

Apply the limit of sum formula:

\int_{0}^{1} \frac{1}{2 + x} d x

Solve

\int_{0}^{1} \frac{1}{2 + x} d x : ln (3) - ln (2)

= ln (3) - ln (2)

\frac{\partial}{\partial x} (5 x^{2} y^{3} + 2 x^{2} + 7 y)

f (x) = e^{- x}, 5

y = ln (5 x \cdot (2 x^{3} - 4 x)^{5})

\int 1 + e^{x} sin (y) d x

\frac{d y}{d x} = \frac{( x + 3 y )}{( y - 3 x )}

What is the limit as n approaches infinity of \sum_{i=1}^n 1/(2n+i) ?
The limit as n approaches infinity of \sum_{i=1}^n 1/(2n+i) is ln(3)-ln(2)