For integers $n,k\geq 1$ we repeatedly toss a coin and count the number of heads that occur. The probabilty of getting a head is $min(t/n,1)$ where $t$ is the current discrete time step. I am trying to work out the asymptotics of the mean time to get $k$ heads. My claim is the following.

Claim:Let $X$ be a random variable which represents the number of coin tosses at the $k$th head. If $k$ is $o(n)$ and $n \rightarrow \infty$, we have: $$\mathbb{E}(X) \sim \sqrt{2n}\frac{\Gamma(k+\frac{1}{2})}{(k-1)!}$$

My "proof" uses a limit approximation of a nonhomogeneous Bernoulli process by a nonhomogeneous Poisson process. We then compute the expected number of tosses by transforming the nonhomogeneous Poission process to a homogeneous process by an application of the inverse transform method.

Let $T_k$ be the time taken to see $k$ events in a Poisson process with increasing rate $t/n$, $t >0$. Define $$\Lambda(t) \overset{\text{def}}{=}\;\; \int_0^t \frac{x}{n} dx = \frac{t^2}{2n}$$ with inverse $$\Lambda^{-1}(y) = \sqrt{2 y n}.$$

The key observation is that $T_k$ has the same distribution as $\Lambda^{-1}(S_k)$, where $S_k$ is the time taken to see $k$ events
in a Poisson process with rate $1$.

We want
\begin{align*}
\mathbb{E}(T_k) &= \mathbb{E}(\sqrt{2n S_k}) \\
&= \sqrt{2n} \int_0^\infty \frac{x^{1/2} x^{k-1} e^{-x}}{(k-1)!} \mathrm{d}x\\
&= \frac{\Gamma(k+\frac{1}{2})}{(k-1)!}.
\end{align*}

The problem is that when $k$ grows with $n$, as is permitted in the claim, I don't see how to give a formal justification for this line of reasoning. In particular, how can you justify the limit approximation?

Any help gratefully received.

As Douglas Zare points out, when $k$ grows with $n$ (and is also $o(n)$), the claim is equivalent to $$\mathbb{E}(X) \sim \sqrt{2nk}.$$

If my particular approach can't be justified, is there another way to get the same result which has a surer footing?