Why is this sequence a [n] unable to correctly find its general term formula?

Question

Given the sequence a [n], a [n] represents the general term of the sequence. Known conditions include:

a[n + 1] == a[n] + Log[n/(n + 1)], a[1] == 1

The result obtained from this is:

ClearAll["`*"]
RSolve[{a[n + 1] == a[n] + Log[n/(n + 1)], a[1] == 1}, a[n], 
  n] // FullSimplify

How to obtain the correct answer：

a[n] == 1 - Log[n]

$\begingroup$ Both expressions give the same sequence. $\endgroup$
– Daniel Huber
Commented Apr 29, 2023 at 15:29 — Daniel Huber, Commented Apr 29, 2023 at 15:29

user64494 · Accepted Answer · 2023-04-29 14:25:15Z

3

This can be done as follows.

FunctionExpand[Log[n/(n + 1)], Assumptions -> n \[Element] PositiveIntegers]

Log[n] - Log[1 + n]

RSolve[{a[n + 1] == a[n] + Log[n] - Log[(n + 1)], a[1] == 1}, a[n], n]

{{a[n] -> 1 - Log[n]}}

answered Apr 29, 2023 at 14:25

user64494

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1

$\begingroup$ Or by FunctionExpand[ 1 + Derivative[1, 0][Zeta][0, n] - Derivative[1, 0][Zeta][0, 1 + n], Assumptions -> n \[Element] PositiveIntegers] which results in 1+Log[Gamma[n]]-Log[Gamma[1+n]]. ` $\endgroup$
– user64494
Commented Apr 29, 2023 at 14:32
2

$\begingroup$ +1 Or put the FunctionExpand inside, e.g., RSolve[{a[n + 1] == a[n] + Log[n/(n + 1)] // FunctionExpand, a[1] == 1}, a[n], n] $\endgroup$
– Bob Hanlon
Commented Apr 29, 2023 at 16:04

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