Find estimates for the solution $x=x_n$ of the equation $(1+x)^n =1+(n+1)x$

Question

It is clear that for every $n> 0$, and $\delta> 0$, we have $(1+x)^n < 1+ (n+\delta) x$, for $x>0$ small. In particular we have $(1+x)^n< 1 + (n+1) x$ for $x> 0$ small enough. How small is just the question. I have approximated the logs on both sides with their quadratic terms and got $x_n\simeq \frac{2}{n^2+n+1}$ for the solution of the equation $$(1+x_n)^n = 1+(n+1) x_n$$ Now it turns out that we have $$x_n > \frac{2}{n^2}$$ equivalently $$\left(1+\frac{2}{n^2}\right)^n < 1+\frac{2(n+1)}{n^2}$$

This is not hard to show with some calculus ( take the second derivative).

I am interested in better estimates for $x_n$ defined above. I am thinking about large $n$'s

Any feedback would be appreciated!

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Got some great feedback.

1. @Somos solution finds it as a power series in $\frac{1}{n}$. It seems that all of the coefficients are positive. I wonder how one could prove that. Also @Somos expressed the problem as an inversion of a function.

2. In @Claude Leibovici solution it seems we find the Pade approximants. They themselves have expansion in $\frac{1}{n}$ with positive coefficients it seems. Intriguing.

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This is still evolving. The way I see it now:

Let $t = \frac{1}{n}$, $t$ small. Also take $k$ a parameter ( assumed positive).

The equation in $x$ with parameters $k$ and $t$

$$(1+ x)^{\frac{1}{t}} - (1 + (\frac{1}{t} + k) x)$$

has a solution for $x$ that is approximately $2 k t^2$. So write $x = 2 k t^2 \cdot y$ and consider the equation in $y$

$$\frac{ \log( 1+ 2 k t^2 y) - t \log (1 + (\frac{1}{t} + k) 2 k t^2 y)}{t^3} =0$$

or $f(k,t,y) = 0$. In this setup $f$ is analytic (in $k$, $t$, $y$), $f(k, 0, 1) = 0$, and moreover, $\frac{\partial f(k,0,1)}{\partial y} = 2 k^2 \ne 0$ (if $k\ne 0$). So now we get a solution

$$x= x(t) = 2 k t^2[ 1+ \frac{1}{3}( 3 - 2k)t + \frac{1}{9} ( 9+ 5 k^2) + \frac{1}{135}( 135 + 90 k - 30 k^2 - 68 k^3)t^3 + \cdots ]$$

This series is convergent for small $t$. For $k=1$ ( the original problem) it seems the coefficients are positive, still not elucidated.

With calculus ( using the known series expansion of @Somos) we could show that

$$x(t) = x(1,t) > 2 t^2( 1 + 1/3 t + 14 /9 t^2)$$

by showing that

$$f(t) \colon = t \log ( 1+ (\frac{1}{t} + 1)2 t^2( 1 + 1/3 t + 14/9 t^2)) - \log ( 1 + 2 t^2( 1 + 1/3 t + 14/9 t^2))>0$$

For this, check that $f(0) = f'(0) = 0$, and $f''(t) > 0$ for $t> 0$ ( see this WA link )

Answer 1

The problem is to find the smallest positive real solution $\,x_n\,$ of

$$ (1+x_n)^n = 1+(n+1) x_n. \tag1 $$

One way is to use the method of undetermined coefficients. Begin with the Ansatz

$$ x_n = a_0 + a_1n^{-1} + a_2n^{-2} + \cdots \tag2 $$

where the coefficients $\,a_0,a_1,a_2,\dots\,$ are to be determined. Easy estimates indicate that $\,a_0 = a_1 = 0, a_2 = 2.\,$

Now to find the next coefficient, let

$$ x_n = 2n^{-2} + a_3n^{-3} + a_4n^{-4} + \cdots. \tag3 $$

So compute

$$ (1+x_n)^n = 1 + 2n^{-1} + (2+a_3)n^{-2} + (-2/3+2a_3+a_4)n^{-3} + \cdots \tag4 $$

and

$$ 1+(n+1)x_n = 1 + 2n^{-1} + (2+a_3)n^{-2} + (a_3+a_4)n^{-3} + \cdots.\tag5 $$

Equate the two series to get $\,a_3 = 2/3.\,$ Repeat this procedure to get

$$ x_n = 2n^{-2} +\frac23n^{-3} + \frac{28}9n^{-4} + \frac{254}{135}n^{-5} + \frac{1886}{405}n^{-6} + \cdots. \tag6 $$

Some more computation strongly suggests that all of the coefficients of the series for $\,x_n\,$ from $\,n^{-2}\,$ on are positive and that the series converges for $\,n>1\,$ since $\,x_n\to\infty\,$ as $\,n\,$ decreases to $1$. The denominators seem to be somehow related to OEIS sequence A141143.

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There is an alternate way to solve a slightly generalized problem. Let

$$ (1+x_n)^n = 1+(n+z) x_n \tag7 $$

where $\,z\,$ is a new variable and $\,z=1\,$ in the original problem. Use the binomial theorem to expand the left side and solve for $\,z\,$ to get

$$ z = {n \choose 2}x_n + {n \choose 3}x_n^2 + \cdots. \tag8 $$

Use power series reversion to express $\,x_n\,$ in terms of $\,z\,$ as

$$ x_n = \frac{2}{n(n-1)}z + \frac{4(2-n)}{3n^2(n-1)^2}z^2 + \frac{2(14-17n+5n^2)}{9n^3(n-1)^3}z^3 + \cdots. \tag9 $$

Expand this as a power series in powers of $\,n^{-1}\,$ to get

$$ x_n = 2zn^{-2} - \frac23(2z^2-3z)n^{-3} +\frac29(5z^3+9z)n^{-4} +\cdots. \tag{10} $$

When $\,z=1\,$ this agrees with the series in equation $(6)$.

Answer 2

What you can also is to consider that you look for the zero of function $$f(x)=\frac 1 x \left((x+1)^n-(n+1) x-1\right)$$ and perform one single iteration of a Newton-like method of order $n$.

Thsi would give $$x_{(2)}=\frac{2}{(n-1) n}$$ $$x_{(3)}=\frac{6}{3 n^2-n-4}$$ $$x_{(4)}=\frac{2 (3 n-4)}{3 n^3-5 n^2-3 n+6}$$ $$x_{(5)}=\frac{30 \left(3 n^3-5 n^2-3 n+6\right)}{45 n^5-90 n^4-85 n^3+216 n^2+40 n-144}$$ $$x_{(6)}=\frac{2 \left(45 n^5-90 n^4-85 n^3+216 n^2+40 n-144\right)}{45 n^7-105 n^6-120 n^5+377 n^4+95 n^3-488 n^2-20 n+240}$$ and so on.

Using $x_{(6)}$ and expanding it for large values of $n$

$$x_{(6)}=\frac{2}{n^2}+\frac{2}{3 n^3}+\frac{28}{9 n^4}+\frac{254}{135 n^5}+\frac{1886}{405 n^6}+O\left(\frac{1}{n^7}\right)$$ which, for sure, is the same as what @Somos gave in his answer.