How Can You Solve for $a_n$ in the Sequence Given $2S_n = a_n + \frac{1}{a_n}$?

[Albert1]

A sequence {$a_n$},$S_n=a_1+a_2+-----+a_n$,for each $n\in N, a_n>0$

if $2S_n=a_n+\dfrac {1}{a_n}$

please find $a_n=?$ (express in $n$)

 

[Greg]

A sequence {$a_n$},$S_n=a_1+a_2+-----+a_n$,for each $n\in N, a_n>0$

if $2S_n=a_n+\dfrac {1}{a_n}$

please find $a_n=?$ (express in $n$)

Spoiler

I claim the sequence is $1,\sqrt2-1,\sqrt3-\sqrt2,\sqrt4-\sqrt3,\dots$ and thus the general term is
$$a_n=\sqrt n-\sqrt{n-1}$$

$$2S_n=a_n+\dfrac{1}{a_n}\implies S_n=\dfrac{a^2_n+1}{2a_n}$$

Proof by induction:

We have
$$1+(\sqrt2-1)+(\sqrt3-\sqrt2)+\dots+\sqrt{n}-\sqrt{n-1}=\sqrt n$$
$$S_{n+1}=\dfrac{(\sqrt{n+1}-\sqrt{n})^2+1}{2(\sqrt{n+1}-\sqrt{n})}$$
$$=\dfrac{n+1-2\sqrt{n+1}\sqrt{n}+n+1}{2(\sqrt{n+1}-\sqrt{n})}$$
$$=\dfrac{2(n+1)-2\sqrt{n+1}\sqrt{n}}{2(\sqrt{n+1}\sqrt{n})}$$
$$=\dfrac{2\sqrt{n+1}(\sqrt{n+1}-\sqrt{n})}{2(\sqrt{n+1}-\sqrt{n})}=\sqrt{n+1}$$

It follows that $a_n=\sqrt{n}-\sqrt{n-1}$.

 

[Albert1]

Spoiler

I claim the sequence is $1,\sqrt2-1,\sqrt3-\sqrt2,\sqrt4-\sqrt3,\dots$ and thus the general term is
$$a_n=\sqrt n-\sqrt{n-1}$$

$$2S_n=a_n+\dfrac{1}{a_n}\implies S_n=\dfrac{a^2_n+1}{2a_n}$$

Proof by induction:

We have
$$1+(\sqrt2-1)+(\sqrt3-\sqrt2)+\dots+\sqrt{n}-\sqrt{n-1}=\sqrt n$$
$$S_{n+1}=\dfrac{(\sqrt{n+1}-\sqrt{n})^2+1}{2(\sqrt{n+1}-\sqrt{n})}$$
$$=\dfrac{n+1-2\sqrt{n+1}\sqrt{n}+n+1}{2(\sqrt{n+1}-\sqrt{n})}$$
$$=\dfrac{2(n+1)-2\sqrt{n+1}\sqrt{n}}{2(\sqrt{n+1}\sqrt{n})}$$
$$=\dfrac{2\sqrt{n+1}(\sqrt{n+1}-\sqrt{n})}{2(\sqrt{n+1}-\sqrt{n})}=\sqrt{n+1}$$

It follows that $a_n=\sqrt{n}-\sqrt{n-1}$.

thanks greg1313:
very good!your answer is correct

Spoiler

$given: \,\, 2S_n=a_n+\dfrac{1}{a_n}---(1)\\$
$for \,\,n=1\\$
$S_1=a_1=1\\$
$for \,\,n\geq 2\\$
$a_n=S_n-S_{n-1}---(2)\\$
$put \,\, (2)\,\, to \,\,(1)\\$
we have :$S_n^2-S_{n-1}^2=1\,\,\therefore S_n^2=1+n-1=n\\$
$\therefore S_n=\sqrt n$
from (2):$a_n=\sqrt{n}-\sqrt{n-1}$ #

 

[MarkFL]

thanks greg1313:
very good!your answer is correct

Spoiler

$given: \,\, 2S_n=a_n+\dfrac{1}{a_n}---(1)\\$
$for \,\,n=1\\$
$S_1=a_1=1\\$
$for \,\,n\geq 2\\$
$a_n=S_n-S_{n-1}---(2)\\$
$put \,\, (2)\,\, to \,\,(1)\\$
we have :$S_n^2-S_{n-1}^2=1\,\,\therefore S_n^2=1+n-1=n\\$
$\therefore S_n=\sqrt n$
from (2):$a_n=\sqrt{n}-\sqrt{n-1}$ #

A few points:

Spoiler

Albert, you seem to assume $a_1=1$ in your proof, or at least did not clearly show this to be true. What I would do here is state:

\(\displaystyle 2S_1=a_1+\frac{1}{a_1}\tag{1}\)

\(\displaystyle S_1=a_1\tag{2}\)

Subtracting (2) from (1), we obtain:

\(\displaystyle S_1=\frac{1}{a_1}\)

Now, substituting for $S_1$ into (2), we have:

\(\displaystyle \frac{1}{a_1}=a_1\implies a_1^2=1\implies S_1^2=1\)

Now, observing that:

\(\displaystyle S_n-S_{n-1}=a_n\), we may write:

\(\displaystyle 2S_n=S_n-S_{n-1}+\frac{1}{S_n-S_{n-1}}\)

from which we obtain:

\(\displaystyle S_n^2-S_{n-1}^2=1\)

And from this, we may write:

\(\displaystyle S_n^2-S_1^2=\sum_{k=2}^n\left(S_{k}^2-S_{k-1}^2\right)=\sum_{k=2}^n\left(1\right)=n-1\)

And from this we get:

\(\displaystyle S_n^2=n\)

Given $0<a_1$, we find:

\(\displaystyle S_n=\sqrt{n}\)

Hence:

\(\displaystyle a_n=\sqrt{n}-\sqrt{n-1}\)