Evaluating Sum Compute: n+1 Roots

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The discussion focuses on evaluating 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 compute the floor of the sum of various roots, specifically from $$n+1$$ down to $$2$$. The conversation highlights the importance of correctly interpreting the problem statement and emphasizes the final result as a floor function of the computed sum.

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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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