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.

  • $\begingroup$ Hint to finish off the first part: show that $1\le a_n\le 1+\frac1n$ for all $n$. $\endgroup$ – 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:


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:


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

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

we get

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

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

First part:

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


$\log z_n$ is decreasing $\rightarrow$

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

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

Bounded below:


Hence convergent.

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


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