Real Analysis/Section 2 Exercises
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- For the following list of sequences determine whether the following sequences converge or diverge directly form the definition of convergence:
- the sequence
; - the sequence
; - the sequence
; - The recursive sequence defined by the following:
- x0 = 2

- the sequence
- Let
.
- For the following sequences determine for which real values x does the given sequence converge, and what the sequence converges to:
;
;
for an arbitrary real number x;
- Given any real number c, find a recursively defined sequence that converges to
. - Given a sequence (xn), and a natural number k, define a sequence yn by yn = xn+k. Show that (xn) is convergent if and only if (yn) is convergent. Show further that when they converge they converge to the same limit.
- Suppose that the sequences (xn) and (yn) converge to a real number a. Show that the sequence (zn) defined by
- Let (xn) be a sequence of real numbers and let (yn) be a sub-sequence. Suppose (yn) is convergent, show that (xn) may not necessarily be convergent.
- Suppose that (xn) is a convergent sequence that does not converge to 0. Further assume that for all n in N, xn ≠ 0. Show that there exists δ > 0 so that |xn| > δ and |lim xn | > δ.
- Cesaro Mean convergence: We say a sequence (xn) converges to x by Cesaro means if the sequence of averages yn = (x1 + x2 + … + xn)/n converges to x. Suppose (xn) converges to a real number x, show that (xn) converges by Cesaro means to x. Give an example to show that a divergent sequence (xn) may converge by Cesaro means.
- Find the the sequence of Cesaro means for (1, 1, -1, 1, 1, -1...) and determine if the converge. If they converge, find the limit.
- Consider the recursively defined sequence given by x1 = 1 and xn = 1 + 1/xn. Show that xn converges and find its limit.
- In our discussion of telescoping series we showed that a telescoping series converged to a1 − lim aN+1 and for this to hold it was not necessary to have that lim aN+1 = 0. Indeed, it is correct that this is not necessary. On the other hand we later proved that for a convergent series the limit of the terms must be 0. How can both by correct? Explain why, in our set up, we may have a convergent telescoping series such that lim aN+1 ≠ 0, but it is still true that for every convergent series the limit of the terms is 0.
- Suppose that cn ≤ an ≤ bn for all natural numbers n. Show that if both ∑ cn and ∑ bn converge, then ∑ an converges.
