We can express f[n]
as a continued fraction:
f[n] == ContinuedFractionK[k, 1, {k, 1, n}]
Now unfortunately, Mathematica says the following diverges, when it really doesn't:
ContinuedFractionK[k, 1, {k, 1, Infinity}]
ContinuedFractionK::div: The continued fraction does not converge. >>
ContinuedFractionK[k,1,{k,1,Infinity}]
We can easily approximate the limit though:
ContinuedFractionK[n, 1., {n, 1, 10000}]
0.525135
Edit: a way to find the closed form of the limit
The value of your desired limit is
$$ \frac{\sqrt{2/(e \pi)}}{\text{erfc}(1/\sqrt{2})} - 1 \approx 0.525135. $$
Here's how I found it.
Since ContinuedFractionK[k, 1, {k, 1, n}]
returns

if we can solve these recurrence relations, we can potentially find the limit.
Now I was unable to find a way to solve these, but what I was able to do was find the closed form of their exponential generating functions. Taking the limit of the quotient of these e.g.f's will give the same result.
Now both recurrences look like
a[n+2] - a[n+1] - n a[n] - 2a[n] == 0
with some initial conditions. If these are Taylor series coefficients to a function f[x]
, then a[n+2]
corresponds to f''[x]
, a[n+1]
corresponds to f'[x]
, n a[n]
corresponds to x f'[x]
, and a[n]
corresponds to f[x]
. This gives us an ODE that f[x]
satisfies. Solving it (with the appropriate initial conditions) will give us our e.g.f's.
(* e.g.f of numerator *)
E1 = DSolveValue[{f''[x] - (x + 1)f'[x] - 2f[x] == 0, f[0] == 0, f'[0] == 1}, f[x], x];
(* e.g.f of denominator *)
E2 = DSolveValue[{f''[x] - (x + 1)f'[x] - 2f[x] == 0, f[0] == 1, f'[0] == 1}, f[x], x];
limit = Limit[E1/E2, x -> Infinity] // FullSimplify
Sqrt[2/(E π)]/Erfc[1/Sqrt[2]] - 1
N[limit]
0.525135