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

  • Thread starter Thread starter Albert1
  • Start date Start date
  • Tags Tags
    Sequence
AI Thread Summary
The discussion focuses on finding the expression for the sequence {$a_n$} given the equation 2S_n = a_n + 1/a_n, where S_n is the sum of the first n terms of the sequence. Participants confirm the correctness of the solution provided by a user, indicating that the problem has been resolved satisfactorily. The sequence is defined for positive values of a_n, and the goal is to express a_n in terms of n. The conversation emphasizes the importance of understanding the relationship between the terms and their cumulative sum. Overall, the problem was successfully addressed with a clear solution.
Albert1
Messages
1,221
Reaction score
0
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 $n$)
 
Mathematics news on Phys.org
Albert said:
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 $n$)

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 $a_n=\sqrt{n}-\sqrt{n-1}$.
 
greg1313 said:
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
$given: \,\, 2S_n=a_n+\dfrac{1}{a_n}---(1)\\$
$for \,\,n=1\\$
$S_1=a_1=1\\$
$for \,\,n\geq 2\\$
$a_n=S_n-S_{n-1}---(2)\\$
$put \,\, (2)\,\, to \,\,(1)\\$
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):$a_n=\sqrt{n}-\sqrt{n-1}$ #
 
Last edited:
Albert said:
thanks greg1313:
very good!your answer is correct
$given: \,\, 2S_n=a_n+\dfrac{1}{a_n}---(1)\\$
$for \,\,n=1\\$
$S_1=a_1=1\\$
$for \,\,n\geq 2\\$
$a_n=S_n-S_{n-1}---(2)\\$
$put \,\, (2)\,\, to \,\,(1)\\$
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:

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:

$$2S_1=a_1+\frac{1}{a_1}\tag{1}$$

$$S_1=a_1\tag{2}$$

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

$$S_1=\frac{1}{a_1}$$

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

$$\frac{1}{a_1}=a_1\implies a_1^2=1\implies S_1^2=1$$

Now, observing that:

$$S_n-S_{n-1}=a_n$$, we may write:

$$2S_n=S_n-S_{n-1}+\frac{1}{S_n-S_{n-1}}$$

from which we obtain:

$$S_n^2-S_{n-1}^2=1$$

And from this, we may write:

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

And from this we get:

$$S_n^2=n$$

Given $0<a_1$, we find:

$$S_n=\sqrt{n}$$

Hence:

$$a_n=\sqrt{n}-\sqrt{n-1}$$
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...

Similar threads

Replies
1
Views
1K
Replies
2
Views
1K
Replies
2
Views
1K
Replies
1
Views
1K
Replies
4
Views
3K
Replies
8
Views
2K
Replies
2
Views
1K
Back
Top