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Given that $\{u_n\}$ is a monotonically increasing bounded sequence, now I want to judge whether the following sequence converges:

$$\begin{array}{c} A. \sum_{n = 1}^{\infty} \frac{u_{n}}{n} & B. \sum_{n = 1}^{\infty}(-1)^{n} \frac{1}{u_{n}} \\ C. \sum_{n = 1}^{\infty}\left(1-\frac{u_{n}}{u_{n+1}}\right) & D. \sum_{n = 1}^{\infty}\left(u_{n+1}^{2}-u_{n}^{2}\right) \end{array}$$

Since $\{u_n\}$ is an abstract sequence, I don't know how to use MMA to solve this problem.

Therefore, I use the following special case to verify the problem:

u[n_] := Sqrt[n/(n + 1)]

Sum[u[n]/n, {n, 1, Infinity}]
Sum[(-1)^n/u[n], {n, 1, Infinity}]
Sum[1 - u[n]/u[n + 1], {n, 1, Infinity}]
Sum[u[n + 1]^2 - u[n]^2, {n, 1, Infinity}]

However, MMA cannot judge whether Sum[1 - u[n]/u[n + 1], {n, 1, Infinity}] converges or not.

How can I judge the convergence of option C correctly?

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1 Answer 1

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Clear["Global`*"]

u[n_] := Sqrt[n/(n + 1)]

SumConvergence indicates that the sum converges.

SumConvergence[1 - u[n]/u[n + 1], n]

(* True *)

Using a specific Method,

Assuming[n > 0, {#, SumConvergence[
      1 - u[n]/u[n + 1] // Simplify, n, Method -> #]} & /@
   {"RatioTest", 
    "RootTest", "RaabeTest", "IntegralTest"}] //
 Grid[#, Frame -> All] &

enter image description here

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  • $\begingroup$ Thank you very much. Do you have a way to deal with abstract sequences directly? $\endgroup$ Jul 29, 2020 at 6:37
  • $\begingroup$ Since summands converge to zero faster than 1/n, the sum converges for n-> Infinity ser = Series[1 - u[n]/u[n + 1], {n, \[Infinity], 2}] // Normal yields 1/(2 n^2) . $\endgroup$
    – Akku14
    Jul 29, 2020 at 8:28

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