Evaluating Sum Compute: n+1 Roots

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Discussion Overview

The discussion revolves around the computation of a specific sum involving roots, expressed as $$\sqrt[n+1]{\frac{n+1}{n}}+\sqrt[n]{\frac{n}{n-1}}+\cdots+\sqrt[4]{\frac{4}{3}}+\sqrt[3]{\frac{3}{2}}+\sqrt{2}$$. The focus is on evaluating this expression and understanding its properties.

Discussion Character

  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • Participants present the expression to be computed, with some clarifying the need to take the floor of the sum.
  • There is a repetition of the clarification regarding the floor function, indicating a potential misunderstanding or emphasis on the problem's requirements.
  • One participant acknowledges another's contribution positively, suggesting engagement with the problem-solving process.

Areas of Agreement / Disagreement

Participants appear to agree on the expression to be computed, but there is a lack of detailed exploration of the computation itself, leaving the discussion somewhat unresolved regarding the actual evaluation.

Contextual Notes

The discussion does not delve into the assumptions or methods for evaluating the sum, nor does it clarify the implications of the floor function on the final result.

anemone
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Compute $$\sqrt[n+1]{\frac{n+1}{n}}+\sqrt[n]{\frac{n}{n-1}}+\cdots+\sqrt[4]{\frac{4}{3}}+\sqrt[3]{\frac{3}{2}}+\sqrt{2}$$.
 
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anemone said:
Compute $$\sqrt[n+1]{\frac{n+1}{n}}+\sqrt[n]{\frac{n}{n-1}}+\cdots+\sqrt[4]{\frac{4}{3}}+\sqrt[3]{\frac{3}{2}}+\sqrt{2}=S$$.
it should be :
for any $n$
$n<S<n+1$
 
Albert said:
it should be :
for any $n$
$n<S<n+1$

Ops! You're right, and the problem is actually asked to compute $$
\left\lfloor{\sqrt[n+1]{\frac{n+1}{n}}+\sqrt[n]{\frac{n}{n-1}}+\cdots+\sqrt[4]{\frac{4}{3}}+\sqrt[3]{\frac{3}{2}}+\sqrt{2}}\right\rfloor$$.:o
 
anemone said:
Ops! You're right, and the problem is actually asked to compute $$
\left\lfloor{\sqrt[n+1]{\frac{n+1}{n}}+\sqrt[n]{\frac{n}{n-1}}+\cdots+\sqrt[4]{\frac{4}{3}}+\sqrt[3]{\frac{3}{2}}+\sqrt{2}}\right\rfloor$$.:o

We have for t positive integer
$(1+x)^{t+1} > 1 + x(t+1)$
or $(1+x) > \sqrt[t+1]{1 + x(t+1)}$
putting $x = \frac{1}{t(t+1)}$ we get
$1+\frac{1}{t(t+1)} > \sqrt[t+1]{1 + \frac{1}{t}}$
or $1+\frac{1}{t(t+1)} > \sqrt[t+1]{\frac{t+1}{t}}$
or $1+\frac{1}{t} - \frac{1}{t+1} > \sqrt[t+1]{\frac{t+1}{t}}$
adding the above from t = 1 to n we get
${\sqrt[n+1]{\frac{n+1}{n}}+\sqrt[n]{\frac{n}{n-1}}+\cdots+\sqrt[4]{\frac{4}{3}}+\sqrt[3]{\frac{3}{2}}+\sqrt{2}} <n + 1 - \frac{1}{n+1}$
as each term is greater than one we have
${\sqrt[n+1]{\frac{n+1}{n}}+\sqrt[n]{\frac{n}{n-1}}+\cdots+\sqrt[4]{\frac{4}{3}}+\sqrt[3]{\frac{3}{2}}+\sqrt{2}} > n $
as the value is between n and n + 1 so
$\left\lfloor{\sqrt[n+1]{\frac{n+1}{n}}+\sqrt[n]{\frac{n}{n-1}}+\cdots+\sqrt[4]{\frac{4}{3}}+\sqrt[3]{\frac{3}{2}}+\sqrt{2}}\right\rfloor = n $
 
kaliprasad said:
We have for t positive integer
$(1+x)^{t+1} > 1 + x(t+1)$
or $(1+x) > \sqrt[t+1]{1 + x(t+1)}$
putting $x = \frac{1}{t(t+1)}$ we get
$1+\frac{1}{t(t+1)} > \sqrt[t+1]{1 + \frac{1}{t}}$
or $1+\frac{1}{t(t+1)} > \sqrt[t+1]{\frac{t+1}{t}}$
or $1+\frac{1}{t} - \frac{1}{t+1} > \sqrt[t+1]{\frac{t+1}{t}}$
adding the above from t = 1 to n we get
${\sqrt[n+1]{\frac{n+1}{n}}+\sqrt[n]{\frac{n}{n-1}}+\cdots+\sqrt[4]{\frac{4}{3}}+\sqrt[3]{\frac{3}{2}}+\sqrt{2}} <n + 1 - \frac{1}{n+1}$
as each term is greater than one we have
${\sqrt[n+1]{\frac{n+1}{n}}+\sqrt[n]{\frac{n}{n-1}}+\cdots+\sqrt[4]{\frac{4}{3}}+\sqrt[3]{\frac{3}{2}}+\sqrt{2}} > n $
as the value is between n and n + 1 so
$\left\lfloor{\sqrt[n+1]{\frac{n+1}{n}}+\sqrt[n]{\frac{n}{n-1}}+\cdots+\sqrt[4]{\frac{4}{3}}+\sqrt[3]{\frac{3}{2}}+\sqrt{2}}\right\rfloor = n $

I privately told anemone that the answer was $n$, but I was told I must show my work. (Crying)
 
kaliprasad said:
We have for t positive integer
$(1+x)^{t+1} > 1 + x(t+1)$
or $(1+x) > \sqrt[t+1]{1 + x(t+1)}$
putting $x = \frac{1}{t(t+1)}$ we get
$1+\frac{1}{t(t+1)} > \sqrt[t+1]{1 + \frac{1}{t}}$
or $1+\frac{1}{t(t+1)} > \sqrt[t+1]{\frac{t+1}{t}}$
or $1+\frac{1}{t} - \frac{1}{t+1} > \sqrt[t+1]{\frac{t+1}{t}}$
adding the above from t = 1 to n we get
${\sqrt[n+1]{\frac{n+1}{n}}+\sqrt[n]{\frac{n}{n-1}}+\cdots+\sqrt[4]{\frac{4}{3}}+\sqrt[3]{\frac{3}{2}}+\sqrt{2}} <n + 1 - \frac{1}{n+1}$
as each term is greater than one we have
${\sqrt[n+1]{\frac{n+1}{n}}+\sqrt[n]{\frac{n}{n-1}}+\cdots+\sqrt[4]{\frac{4}{3}}+\sqrt[3]{\frac{3}{2}}+\sqrt{2}} > n $
as the value is between n and n + 1 so
$\left\lfloor{\sqrt[n+1]{\frac{n+1}{n}}+\sqrt[n]{\frac{n}{n-1}}+\cdots+\sqrt[4]{\frac{4}{3}}+\sqrt[3]{\frac{3}{2}}+\sqrt{2}}\right\rfloor = n $

Very well done, kaliprasad!(Cool)
 

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