Step by step solution to "Evaluate:limn→∞∑k=1n1n+k2+n"

haugmbd

Answered question

2022-02-23

Evaluate:

lim_{n \to \infty} \sum_{k = 1}^{n} \frac{1}{n + \sqrt{k^{2} + n}}

Ayda Cannon

Beginner2022-02-24Added 3 answers

lim_{n \to \infty} \frac{1}{n} \sum_{k = 1}^{n} \frac{1}{1 + \sqrt{{(\frac{k}{n})}^{2} + \frac{1}{n}}} = lim_{n \to \infty} lim_{m \to \infty} \frac{1}{n} \sum_{k = 1}^{n} \frac{1}{1 + \sqrt{{(\frac{k}{n})}^{2} + \frac{1}{m}}}

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

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

= \log 2

The relevant theorem is if

x_{n m} \to y_{n}

unformly as

m \to \infty

and

x_{n m}

converges as

n \to \infty

, then the double limit exists and

lim_{n \to \infty} x_{\cap} = lim_{n \to \infty} lim_{m \to \infty} x_{n m} = lim_{m \to \infty} lim_{n \to \infty} x_{n m}

To show uniform convergence in this case, note that

| \frac{1}{n} \sum_{k = 1}^{n} \frac{1}{1 + \sqrt{{(\frac{k}{n})}^{2} + \frac{1}{m}}} - \frac{1}{n} \sum_{k = 1}^{n} \frac{1}{1 + \frac{k}{n}} | \leq \frac{1}{n} \sum_{k = 1}^{n} \frac{∣ 1 + \frac{k}{n} - (1 + \sqrt{{(\frac{k}{n})}^{2} + \frac{1}{m}}} {| 1 + \frac{k}{n} | | 1 + \sqrt{{(\frac{k}{n})}^{2} + \frac{1}{m}} |}}{}

\leq \frac{1}{n} \sum_{k = 1}^{n} (\sqrt{{(\frac{k}{n})}^{2} - \frac{1}{m}} - \frac{k}{n})

This is helpful 0 Flag

Randall Odom

Beginner2022-02-25Added 5 answers

Squeezing:

\frac{1}{1 + \sqrt{{(\frac{k}{n})}^{2} + \frac{1}{n}}} < \frac{1}{1 + \sqrt{{(\frac{k}{n})}^{2} + 0}} = \frac{1}{1 + \frac{k}{n}}

And it is obvious that:

lim_{n \to \infty} \frac{1}{n} \sum_{k = 1}^{n} \frac{1}{1 + \frac{k}{n}} = \int_{0}^{1} \frac{1}{1 + x} d x = \ln (2)

We then notice that:

\frac{1}{1 + \sqrt{{(\frac{k}{n})}^{2} + \frac{1}{n}}} > \frac{1}{1 + \frac{k}{n} + \frac{1}{2 k}} > \frac{1 - \frac{1}{2 k}}{1 + \frac{k}{n}} > \frac{1}{1 + \frac{k}{n}} - \frac{1}{2 k}

And finally notice that:

lim_{n \to \infty} \frac{1}{n} \sum_{k = 1}^{n} \frac{1}{1 + \frac{k}{n}} - \frac{1}{2 k} \geq \ln (2)

which follows easily since

\frac{1}{n} \sum_{k = 2}^{n} \frac{1}{k} < \frac{1}{n} \int_{1}^{n} \frac{1}{x} d x = \frac{\ln n}{n}

and this limits may be taken care of by LHospitals rule.

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?