# Find equation for nth value.

1. May 3, 2014

### m84uily

Sn = ∑ni = 1 (Sn - i)-1
S0 = 1

I want to know how to find the general equation for Sn (An example of what I mean by "general equation" would be
Sn = ∑ni = 1i = n(n+1)/2).
Here's S0 though S5:

S0 = 1
S1 = 1
S2 = 1 + 1
S3 = 1 + 1 + 1/2
S4= 1 + 1 + 1/2 + 2/5
S5 = 1 + 1 + 1/2 + 2/5 + 10/29

Thanks in advance, sorry if anything is unclear.

2. May 3, 2014

### The_Duck

I dunno about an exact formula, but it seems that for large $n$, $S_n \approx \sqrt{2 n}$.

3. May 3, 2014

### m84uily

That's neat! How did you get that result?

4. May 3, 2014

### jbunniii

We can view the sum as being a Riemann sum (using rectangles of width 1) corresponding to the following integral equation:
$$f(x) = \int_{0}^{x}\frac{dt}{f(t)}$$
Differentiating both sides, we get
$$f'(x) = \frac{1}{f(x)}$$
By the product rule, this is equivalent to
$$\frac{d}{dx}(f(x)\cdot f(x)) = 2$$
Thus
$$(f(x))^2 = 2x + c$$
and
$$f(x) = \sqrt{2x + c}$$
The integral equation evaluated at $x=0$ forces $c=0$.

So $f(x) = \sqrt{2x}$ is a solution to the integral equation, and therefore, $S_n = \sqrt{2n}$ is an approximate solution to the original problem, when $n$ is large.

5. May 3, 2014

### m84uily

Thanks a lot for explaining that, there's something I don't understand though:
The integral equation evaluated at $x=0$ forces $c=0$.
Why is c=0 forced? Why isn't it c=1?

6. May 3, 2014

### jbunniii

We have
$$f(x) = \int_{0}^{x}\frac{dt}{f(t)}$$
Evaluating this at $x=0$ gives us
$$f(0) = \int_{0}^{0}\frac{dt}{f(t)}$$
The right hand side is an integral over an interval of zero width, so the result is zero. Note that this forces the integrand to have zero denominator at $t=0$, but that's OK: the integral from 0 to 0 is still 0, and $1/\sqrt{2t}$ is (improperly) integrable over $[0,x]$ despite the singularity at $t=0$.

Don't be concerned that $f(0) = 0$ does not match $S_0 = 1$ - no one is claiming that $\sqrt{2n}$ is an exact fit, just that it is a good approximation for large $n$. Note that even if we had $c=1$, the difference between $\sqrt{2n+1}$ and $\sqrt{2n}$ is negligible for large $n$.

7. May 3, 2014

### m84uily

Okay thanks again, I understand now.