# Limits with $a_n+\log a_n = 1+\frac{1}{n}$

If $$a_n+\log a_n = 1+\dfrac{1}{n}$$, compute:

• $$\lim_\limits{n\to \infty}a_n$$
• $$\lim_\limits{n\to \infty} n(a_n-1)$$

If $$a_n$$ is convergent to $$l$$, then passing to limit $$l+\log l = 1$$. But $$f(x)=x+\log x$$ is increasing, so unique solution is $$l = 1$$. But I don't know how to prove $$a_n$$ is bounded and monotonic.

• Hint to finish off the first part: show that $1\le a_n\le 1+\frac1n$ for all $n$. – Greg Martin Feb 10 '20 at 17:57

Well, if you've already seen that $$f:(0,\infty) \to \mathbb{R},\ f(x) = x+\ln x$$ is increasing, then it's pretty clear that the sequence is decreasing, because:

$$f(a_n)=1+\frac{1}{n}>1+\frac{1}{n+1}=f(a_{n+1})$$

However, I think it's easier if you note that $$f$$ is continuous and increasing and that:

$$f(1)=1+\ln 1=1 < 1+\frac{1}{n}$$

$$f\left(1+\frac{1}{n}\right)=1+\frac{1}{n}+\ln\left(1+\frac{1}{n}\right) > 1+\frac{1}{n}$$

so there exists a unique real number $$a_n \in\left(1,1+\frac{1}{n}\right)$$ such that

$$f(a_n) = 1+\frac{1}{n}$$

Now, squeezing, we obviously get $$\lim\limits_{n\to \infty} a_n=1$$.

For the second limit, let $$b_n:=a_n-1\to 0$$ and write the formula as:

$$nb_n\left(1+\frac{\ln(1+b_n)}{b_n}\right)=1$$

Now, just passing to limit and using the well-know:

$$\lim_{x\to 0} \frac{\ln(1+x)}{x}=1$$

we get

$$\lim_{n\to \infty}nb_n=\frac{1}{2}$$

• Your approach via squeezing is really very smart. +1 – Paramanand Singh Feb 12 '20 at 9:05

First part:

$$\log a_ne^{a_n}=1+1/n$$;

$$z_n:=a_ne^{a_n}>e$$;

$$\log z_n$$ is decreasing $$\rightarrow$$

$$z_n$$ is decreasing $$\rightarrow$$ $$a_n$$ is decreasing.

(Note: $$f(x)=xe^x$$ is increasing)

Bounded below:

$$a_n>e/e^{a_n}>0$$.

Hence convergent.

Only solution to $$ae^a =e$$ is $$a=1$$.