Question: "Find the limit: limn→∞(1n+1+1n+2+...+12n)"

Trent Carpenter

Answered question

2020-10-27

Find the limit:

lim_{n \to \infty} (\frac{1}{n + 1} + \frac{1}{n + 2} + . . . + \frac{1}{2 n})

oppturf

Skilled2020-10-28Added 94 answers

Consider the given limit,

lim_{n \to \infty} (\frac{1}{n + 1} + \frac{1}{n + 2} + . . . + \frac{1}{2 n})

now write the limit in the summation form,

lim_{n \to \infty} (\frac{1}{n + 1} + \frac{1}{n + 2} + . . . + \frac{1}{2 n}) = lim_{n \to \infty} \sum_{r = 1}^{n} (\frac{1}{n + r})

= lim_{n \to \infty} \sum_{r = 1}^{n} \frac{1}{n} \cdot (\frac{1}{1 + \frac{r}{n}})

now, convert it into definite integral as follows,
substitute

\frac{r}{n} = x \Rightarrow d x = \frac{1}{n}

so the limits of integration will be

r \to 1, x \to 0

r \to n, x \to 1

so,

lim_{n \to \infty} \sum_{r = 1}^{n} \frac{1}{n} \cdot (\frac{1}{1 + \frac{r}{n}}) = \int_{0}^{1} \frac{1}{1 + x} d x

= [\ln (1 + x)]_{0}^{1}

= (\ln (2) - \ln (1))

= \ln (2)

hence the value of the limit will be ln(2).

Jeffrey Jordon

Expert2022-06-04Added 2605 answers

Answer is given below (on video)

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