A property of positive convergent series Let $(a_k)_{k\geq 1}$ be a positive sequence such that the series $\sum_{k\geq 1} a_k$ is convergent to $L$. Then, by C-S inequality, for any $n\geq 1$, 
$$f(n):=\sum_{k=1}^n \frac{1}{a_k}\geq \frac{n^2}{\sum_{k=1}^n a_k}&gt; \frac{n^2}{L}.$$
Is it true that eventually $f(n)&gt;n^2\ln(n)$ (i.e. there is $n_0$ such that the inequality holds for all $n\geq n_0$).
In mathematical jargon "eventually" would mean the bound holds for all sufficiently large $n$ i.e.
$$\text{there exists $N$ such that $f(n) &gt;n^2\ln n$ for all $n\geq N$.}$$

This is not the case, in fact we can have $f(n)\leq n^2\omega(n)$ infinitely often for any given function $\omega(n)\to\infty$ (in particular for $\omega(n)=\ln n$). This is the best possible because $\liminf f(n)/n^2=+\infty$.

If you instead ask whether $f(n)$ ever exceeds $n^2\ln n$, then the answer is true (even excepting $n=1$). I give two proofs below.

Sequence with $f(n)\leq n^2\omega(n)$ infinitely often given a function $\omega(n)\to\infty$

Let $1\leq n_1\leq n_2\leq \cdots$ be a sequence to be chosen later. Define
$$a_n=\sum_{k\text{ such that }n_k\geq n} \frac{1}{n_k2^k}.$$

Note $\sum_{k\geq 1}a_k$ converges (to $2$), and for each $k\geq 1$,

$$f(n_k)=f(n_{k-1})+\sum_{i=n_{k-1}+1}^{n_k} \frac{1}{a_i} = f(n_{k-1})+2^kn_k(n_k-n_{k-1}).$$

If $n_1,\dots,n_{k-1}$ are fixed, the right-hand side is $O(n_k^2)$. So if $\omega(n_k)\to\infty$ as $n_k\to\infty$, we can always pick $n_k$ large enough such that $f(n_k)\leq (n_k)^2\omega(n_k)$.

Proof of $\liminf f(n)/n^2=+\infty$

Your Cauchy-Schwarz argument similarly gives, for even $n$,

$$f(n)/(n/2)^2&gt;(1/(n/2)^2)\sum_{k=n/2+1}^{n}\frac{1}{a_k} \geq \tfrac 1 4 \left(\sum_{k=n/2+1}^{\infty}a_k\right)^{-1}$$

and the right-hand-side converges to $+\infty$ as $n\to\infty$. Of course odd $n$ obey a similar bound but it is more fiddly to write down.

Grouping proof that $\sup_{n\geq 2} f(n)/n^2\ln n = +\infty$

There must be some $d\geq 1$ such that
$$\sum_{k=2^{d-1}+1}^{2^d}a_k \leq \frac 1{4C \ln(2^d)}$$
since the sum over $d$ of the right-hand-side diverges.

Your Cauchy-Schwarz argument gives
$$f(2^d)&gt;\sum_{k=2^{d-1}+1}^{2^d}\frac{1}{a_k} \geq 4C\ln(2^d)(2^d-2^{d-1})^2=C\ln(2^d)(2^d)^2$$

so $f(n)&gt;Cn^2\ln n$ for $n=2^d$.

Majorization proof that $\sup_{n\geq 2} f(n)/n^2\ln n = +\infty$

Fix a constant C. Define $(r_n)_{n\geq 1}$ to be the unique sequence with
$$\sum_{k=1}^n r_k = R(n) := Cn^2\max(1, \ln n).$$

We can estimate $r_n=O(n\ln n)$ so $\sum 1/r_k$ diverges. Pick $N$ large enough that $\sum_{k=1}^N 1/r_k&gt;L$. Given any sequence $a_1,\dots,a_N&gt;0$ such that $f(n)\leq R(n)$ for $1\leq n\leq N$, we can reduce arbitrary elements of $a_n$ if necessary to increase $f(N)$ to $R(N)$. Then $1/a_1,\dots,1/a_N$ majorizes $r_1,\dots,r_N$: assuming w.l.o.g. that $a_n$ are in descending order,
$$f(n)\leq R(n) \text{ for all $1\leq n&lt;N$, and }f(N)=R(N)$$
(This is majorization "from above", which is equivalent to majorization "from below" because the total sums match.)
Since $1/x$ is a convex function for $x&gt;0$, by Karamata's inequality we have
$$a_1+\dots+a_N\geq \frac{1}{r_1}+\dots+\frac{1}{r_N}$$
which gives $\sum_{k=1}^Na_k &gt;L.$
In mathematical jargon "eventually" would mean the bound holds for all sufficiently large $n$ i.e.
$$\text{there exists $N$ such that $f(n) &gt;n^2\ln n$ for all $n\geq N$.}$$

This is not the case, in fact we can have $f(n)\leq n^2\omega(n)$ infinitely often for any given function $\omega(n)\to\infty$ (in particular for $\omega(n)=\ln n$). This is the best possible because $\liminf f(n)/n^2=+\infty$.

If you instead ask whether $f(n)$ ever exceeds $n^2\ln n$, then the answer is true (even excepting $n=1$). I give two proofs below.

Sequence with $f(n)\leq n^2\omega(n)$ infinitely often given a function $\omega(n)\to\infty$

Let $1\leq n_1\leq n_2\leq \cdots$ be a sequence to be chosen later. Define
$$a_n=\sum_{k\text{ such that }n_k\geq n} \frac{1}{n_k2^k}.$$

Note $\sum_{k\geq 1}a_k$ converges (to $2$), and for each $k\geq 1$,

$$f(n_k)=f(n_{k-1})+\sum_{i=n_{k-1}+1}^{n_k} \frac{1}{a_i} = f(n_{k-1})+2^kn_k(n_k-n_{k-1}).$$

If $n_1,\dots,n_{k-1}$ are fixed, the right-hand side is $O(n_k^2)$. So if $\omega(n_k)\to\infty$ as $n_k\to\infty$, we can always pick $n_k$ large enough such that $f(n_k)\leq (n_k)^2\omega(n_k)$.

Proof of $\liminf f(n)/n^2=+\infty$

Your Cauchy-Schwarz argument similarly gives, for even $n$,

$$f(n)/(n/2)^2&gt;(1/(n/2)^2)\sum_{k=n/2+1}^{n}\frac{1}{a_k} \geq \tfrac 1 4 \left(\sum_{k=n/2+1}^{\infty}a_k\right)^{-1}$$

and the right-hand-side converges to $+\infty$ as $n\to\infty$. Of course odd $n$ obey a similar bound but it is more fiddly to write down.

Grouping proof that $\sup_{n\geq 2} f(n)/n^2\ln n = +\infty$

There must be some $d\geq 1$ such that
$$\sum_{k=2^{d-1}+1}^{2^d}a_k \leq \frac 1{4C \ln(2^d)}$$
since the sum over $d$ of the right-hand-side diverges.

Your Cauchy-Schwarz argument gives
$$f(2^d)&gt;\sum_{k=2^{d-1}+1}^{2^d}\frac{1}{a_k} \geq 4C\ln(2^d)(2^d-2^{d-1})^2=C\ln(2^d)(2^d)^2$$

so $f(n)&gt;Cn^2\ln n$ for $n=2^d$.

Majorization proof that $\sup_{n\geq 2} f(n)/n^2\ln n = +\infty$

Fix a constant C. Define $(r_n)_{n\geq 1}$ to be the unique sequence with
$$\sum_{k=1}^n r_k = R(n) := Cn^2\max(1, \ln n).$$

We can estimate $r_n=O(n\ln n)$ so $\sum 1/r_k$ diverges. Pick $N$ large enough that $\sum_{k=1}^N 1/r_k&gt;L$. Given any sequence $a_1,\dots,a_N&gt;0$ such that $f(n)\leq R(n)$ for $1\leq n\leq N$, we can reduce arbitrary elements of $a_n$ if necessary to increase $f(N)$ to $R(N)$. Then $1/a_1,\dots,1/a_N$ majorizes $r_1,\dots,r_N$: assuming w.l.o.g. that $a_n$ are in descending order,
$$f(n)\leq R(n) \text{ for all $1\leq n&lt;N$, and }f(N)=R(N)$$
(This is majorization "from above", which is equivalent to majorization "from below" because the total sums match.)
Since $1/x$ is a convex function for $x&gt;0$, by Karamata's inequality we have
$$a_1+\dots+a_N\geq \frac{1}{r_1}+\dots+\frac{1}{r_N}$$
which gives $\sum_{k=1}^Na_k &gt;L.$