Limit of Recursive Sequence $y_{n+1}=y_n+\frac{1}{ny_n}$ with $y_1=1$ I came up with this sequence as I was playing around with another question on this site. And so I decided to ask the users if they can find a solution.
Let $$y_1=1$$
$$y_{n+1}=y_n+\frac{1}{ny_n}.$$
Evaluate $$\lim\limits_{n\to \infty} y_n \leq \infty.$$ 
 A: The sequence diverges.  
Suppose the sequence was bounded by some constant, i.e. $y_n < C$ for all $n$. Then in particular $\frac{1}{y_n} > \frac{1}{C}$ for all $n$ so:
$y_{2}>y_1 + \frac{1}{C}$
$y_{3} > y_2 + \frac{1}{2C} > y_1 + \frac{1}{C} + \frac{1}{2C}$
$y_{4} > y_3 + \frac{1}{3C} > y_1 + \frac{1}{C} + \frac{1}{2C} + \frac{1}
{3C}$
...
$y_{n+1} > y_1 + \frac{1}{C} \sum_{i=1}^n \frac{1}{i}$
But the harmonic series diverges.
EDIT:
I'd like to expand a bit on the exact asymptotics in response to Did's comment:
Using the same trick as above but with the squares:
$y_2^2 = y_1^2+\frac{2}{1}+\frac{1}{y_1^2}$
$y_3^2 = y_2^2+\frac{2}{2}+\frac{1}{4y_2^2} = y_1^2+(\frac{2}{1}+\frac{2}{2}) +(\frac{1}{y_1^2} + \frac{1}{4y_2^2})$
... (i.e. by induction)
$y_{n+1} = y_1^2 + 2(\sum_{i=1}^n\frac{1}{i}) + (\sum_{i=1}^n \frac{1}{i^2y_i^2})$
$2H_n + 1 < y_{n+1}^2 < 2H_n + 1 + \frac{\pi^2}{6}$
Where in the last step we use that $y_n \ge 1$ for all $n$ and the sum of the reciprocals of squares sums to $\frac{\pi^2}{6}$.  So indeed $y_n \approx \sqrt{2H_n}$, as suggested.
