- #1

Hall

- 351

- 88

- Homework Statement
- We have to discuss the convergence of ##\sum_{n=1}^{\infty} [\sqrt{n+1} - \sqrt{n}\,]##.

- Relevant Equations
- Cauchy's Criterion says, for the sequence of partial sums ##(s_n)##, if for every ##\varepsilon \lt 0## there exits ##N##, such that for ##m, n \gt N##

$$

| s_n - s_m| \lt \varepsilon

$$

then the series converges.

We're given the series ##\sum_{n=1}^{\infty} [ \sqrt{n+1} - \sqrt{n} ]##.

##s_n = \sqrt{n+1} - 1##

##s_n## is, of course, an increasing sequence, and unbounded, given any ##M \gt 0##, we have ##N = M^2 +2M## such that ##n \gt N \implies s_n \gt M##. Thus, the series must be divergent.

But ##(s_n)## is a Cauchy sequence too, for ##m= n+k##, we have

## |s_{n+k} - s_n| = \sqrt{ n+k+1} - \sqrt{n+1}##

## | s_{n+k} - s_n| = \frac{k}{ \sqrt{n+k+1} + \sqrt{n+1} }##

By increasing ##n## we can make the above difference as small as we please. Thus, the series must convergent.

I do hereby request you gentlemen to offer your advices.

∑n=1∞[n+1−n]

##s_n = \sqrt{n+1} - 1##

##s_n## is, of course, an increasing sequence, and unbounded, given any ##M \gt 0##, we have ##N = M^2 +2M## such that ##n \gt N \implies s_n \gt M##. Thus, the series must be divergent.

But ##(s_n)## is a Cauchy sequence too, for ##m= n+k##, we have

## |s_{n+k} - s_n| = \sqrt{ n+k+1} - \sqrt{n+1}##

## | s_{n+k} - s_n| = \frac{k}{ \sqrt{n+k+1} + \sqrt{n+1} }##

By increasing ##n## we can make the above difference as small as we please. Thus, the series must convergent.

I do hereby request you gentlemen to offer your advices.

∑n=1∞[n+1−n]

Last edited by a moderator: