MHB Evaluating Sum Compute: n+1 Roots

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The discussion focuses on computing the expression $$\left\lfloor{\sqrt[n+1]{\frac{n+1}{n}}+\sqrt[n]{\frac{n}{n-1}}+\cdots+\sqrt[4]{\frac{4}{3}}+\sqrt[3]{\frac{3}{2}}+\sqrt{2}}\right\rfloor$$. Participants clarify that the goal is to find the floor of the sum of these roots. The conversation highlights the importance of correctly interpreting the mathematical notation and the implications for the final result. There is acknowledgment of a previous misunderstanding regarding the problem's requirements. The thread emphasizes the significance of precision in mathematical computations.
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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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