Question

How to prove that $x_n = nq^n$ for $|q| < 1$ is bounded?

score 725 · Answer 1 · 2018-10-29 19:13:46

725

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How to prove that $x_n = nq^n$ for $|q| < 1$ is bounded?

Published on 29 Oct 2018 - 19:13

score 349 · Answer 2 · 2026-03-25 08:05:59

349

Views

Rademacher Complexity Result

Published on 25 Mar 2026 - 8:05

score 95 · Answer 3 · 2018-10-30 11:41:40

95

Views

Upper bound $f(n) \leq c\alpha^n$ with recurrence $f(n) = 2 \cdot f(n-1) + f(n-2)$ with $f(1) = 1$ and $f(2) = 2$

Published on 30 Oct 2018 - 11:41

score 90 · Answer 4 · 2018-10-30 16:57:26

90

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Proof verification of $\sum_{k=1}^n kq^{n-k}$ is unbounded for $n \in \mathbb N$ and $q\in\{\mathbb R\setminus0\}$

Published on 30 Oct 2018 - 16:57

score 160 · Answer 5 · 2018-10-30 18:37:36

160

Views

How to show that $x_{n+1} = \frac{x_n^4 + 1}{5x_n}$ bounded below and above by $1\over 5$ and $2$

Published on 30 Oct 2018 - 18:37

score 105 · Answer 6 · 2018-10-30 20:38:48

105

Views

Concrete mathematics exercise 1.17 upper bound intuition

Published on 30 Oct 2018 - 20:38

score 41 · Answer 7 · 2026-03-28 12:13:25

41

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Showing $2^x$ is a lower bound

Published on 28 Mar 2026 - 12:13

score 129 · Answer 8 · 2018-10-31 15:45:53

129

Views

Show that recurrence relation $x_{n+1} = {1\over 2}\left(x_n + {b^2 \over x_n}\right)$ where $x_1 = a >0$ is bounded.

Published on 31 Oct 2018 - 15:45

score 226 · Answer 9 · 2026-03-28 12:14:47

226

Views

Assuming $f(n)$ is non-decreasing and $f(n) = n^2$ where $n$ is odd, then $f(n) \in \Theta(n^2)$.

Published on 28 Mar 2026 - 12:14

score 705 · Answer 10 · 2026-03-27 23:57:26

705

Views

Bounds on $\binom{2n}{n}$

Published on 27 Mar 2026 - 23:57

score 185 · Answer 11 · 2018-11-01 08:37:48

185

Views

Show that $x_{n+2} = x_{n+1} + \frac{x_n}{2^n}$ is a bounded sequence

Published on 01 Nov 2018 - 8:37

score 140 · Answer 12 · 2018-11-01 13:38:35

140

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Other ways to show $y_n = \prod_{k=1}^n \left(1+{1\over x_k}\right)$ is bounded, where $x_n$ is another sequence with given properties.

Published on 01 Nov 2018 - 13:38

score 92 · Answer 13 · 2018-11-01 22:09:15

92

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How to bound this function?

Published on 01 Nov 2018 - 22:09

score 44 · Answer 14 · 2018-11-02 10:57:10

44

Views

Proof verification of if $S_n = \sum_{k=1}^n x_n$ is bounded then $\{x_n\}$ is bounded

Published on 02 Nov 2018 - 10:57

score 129 · Answer 15 · 2018-11-02 15:28:58

129

Views

Show that $x_n = \sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} - \sqrt n$ is a bounded sequence.

Published on 02 Nov 2018 - 15:28

List Question

How to prove that $x_n = nq^n$ for $|q| < 1$ is bounded?

Rademacher Complexity Result

Upper bound $f(n) \leq c\alpha^n$ with recurrence $f(n) = 2 \cdot f(n-1) + f(n-2)$ with $f(1) = 1$ and $f(2) = 2$

Proof verification of $\sum_{k=1}^n kq^{n-k}$ is unbounded for $n \in \mathbb N$ and $q\in\{\mathbb R\setminus0\}$

How to show that $x_{n+1} = \frac{x_n^4 + 1}{5x_n}$ bounded below and above by $1\over 5$ and $2$

Concrete mathematics exercise 1.17 upper bound intuition

Showing $2^x$ is a lower bound

Show that recurrence relation $x_{n+1} = {1\over 2}\left(x_n + {b^2 \over x_n}\right)$ where $x_1 = a >0$ is bounded.

Assuming $f(n)$ is non-decreasing and $f(n) = n^2$ where $n$ is odd, then $f(n) \in \Theta(n^2)$.

Bounds on $\binom{2n}{n}$

Show that $x_{n+2} = x_{n+1} + \frac{x_n}{2^n}$ is a bounded sequence

Other ways to show $y_n = \prod_{k=1}^n \left(1+{1\over x_k}\right)$ is bounded, where $x_n$ is another sequence with given properties.

How to bound this function?

Proof verification of if $S_n = \sum_{k=1}^n x_n$ is bounded then $\{x_n\}$ is bounded

Show that $x_n = \sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} - \sqrt n$ is a bounded sequence.

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