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Let me try one more time, inspired by Pietro's idea.

Let's for now focus on $t=1/N$ and an interval $I_N$ of length $N$ located near $k\simeq (N/3)\log N$. Let's take $\lambda_k=(1\pm\delta)k$ on this interval, with $\delta=1/\log N$, and the sign is the same as that of $c_k$. Then the terms of $$ \sum_{k\in I_N} c_k(e^{-\lambda_k t}- e^{-kt}) = \sum e^{-k/N}c_k(e^{\pm\delta k/N}-1) \quad\quad\quad\quad (1) $$ are non-negative. In fact, let's also take $c_k$ with alternating signs. Then the sum is $\gtrsim N^{-1/3} \sum |c_k|$. We can easily make this large, for example by giving $|c|$ the constant value $|c_k|=N^{-1/2-\epsilon}$ on $I_N$ (note that this will keep $\sum_{I_N} |c_k|^2$ small, as required by the $\ell^2$ condition).

Now we define the whole sequence by first choosing $N_j$'s that increase very rapidly, and then defining $c_k$ as above on each of these intervals, and $c_k=0 $ otherwise. Notice that for $t=1/N_j$ and $k$ taken from one of the other intervals $I_m$, $m\not= j$, then $e^{-\lambda_kt}-e^{-kt}$ is extremely small, either because $tk$ is either very small or very large. So these intervals make negligible contributions to (1).

It follows that (1) does not go to zero as $t\to 0+$ along the sequence $t=1/N_j$, and as discussed earlier, this means that we have a counterexample.

Finally, an obvious modification also gives examples where $c\in\ell^p$ for any (or all) $p>1$.

Let me try one more time, inspired by Pietro's idea.

Let's for now focus on $t=1/N$ and an interval $I_N$ of length $N$ located near $k\simeq (N/3)\log N$. Let's take $\lambda_k=(1\pm\delta)k$ on this interval, with $\delta=1/\log N$, and the sign is the same as that of $c_k$. Then the terms of $$ \sum_{k\in I_N} c_k(e^{-\lambda_k t}- e^{-kt}) = \sum e^{-k/N}c_k(e^{\pm\delta k/N}-1) \quad\quad\quad\quad (1) $$ are non-negative. In fact, let's also take $c_k$ with alternating signs. Then the sum is $\gtrsim N^{-1/3} \sum |c_k|$. We can easily make this large, for example by giving $|c|$ the constant value $|c_k|=N^{-1/2-\epsilon}$ on $I_N$ (note that this will keep $\sum_{I_N} |c_k|^2$ small, as required by the $\ell^2$ condition).

Now we define the whole sequence by first choosing $N_j$'s that increase very rapidly, and then defining $c_k$ as above on each of these intervals, and $c_k=0 $ otherwise. Notice that for $t=1/N_j$, if $k$ is taken from one of the other intervals $I_m$, $m\not= j$, then $e^{-\lambda_kt}-e^{-kt}$ is extremely small, because $tk$ is either very small or very large. So these intervals make negligible contributions to (1).

It follows that (1) does not go to zero as $t\to 0+$ along the sequence $t=1/N_j$, and as discussed earlier, this means that we have a counterexample.

Finally, an obvious modification also gives examples where $c\in\ell^p$ for any (or all) $p>1$.

The problem with the example below is that $\sum c_k$ doesn't converge, and anyway it would contradict Abel's theorem on power series.

This is not true. Let's take $\lambda_k=k$.

Let's focus on $t=1/N$ and $N\le k\le N\log N$. On the first half of this interval, let's take $c_k=\delta N^{-1/2}$, with a small $\delta>0$ to be determined later, and $c_k=-\delta N^{-1/2}$ on the second half of this interval. Then $$ \sum_{k=N}^{N\log N} e^{-k/N}c_k = e^{-1}\delta N^{1/2} + O(\delta) , $$ which of course is a disaster since $\sum_{k=N}^{N\log N} c_k = 0$.

We now just take a very rapidly increasing sequence $N_j$ and $\delta_j=1/j$, say, and then define $c_k$ as above on the corresponding intervals and $c_k=0$ otherwise. Then for $t=1/N_j$, we have $\sum e^{-kt} c_k \simeq N_j^{1/2}/j\not\to 0=\sum c_k$: for $N_j\le k\le N_j\log N_j$, the discussion above applies, and for the other intervals, $e^{-k/N_j}$ is either very close to $1$ or very small, so these sums are extremely close to zero.

Finally, note that $c\in\ell^2$, as desired; in fact, an obvious modification gives us examples with $c\in\ell^p$ for any (or all) $p>1$.

Let me try one more time, inspired by Pietro's idea.

Let's for now focus on $t=1/N$ and an interval $I_N$ of length $N$ located near $k\simeq (N/3)\log N$. Let's take $\lambda_k=(1\pm\delta)k$ on this interval, with $\delta=1/\log N$, and the sign is the same as that of $c_k$. Then the terms of $$ \sum_{k\in I_N} c_k(e^{-\lambda_k t}- e^{-kt}) = \sum e^{-k/N}c_k(e^{\pm\delta k/N}-1) \quad\quad\quad\quad (1) $$ are non-negative. In fact, let's also take $c_k$ with alternating signs. Then the sum is $\gtrsim N^{-1/3} \sum |c_k|$. We can easily make this large, for example by giving $|c|$ the constant value $|c_k|=N^{-1/2-\epsilon}$ on $I_N$ (note that this will keep $\sum_{I_N} |c_k|^2$ small, as required by the $\ell^2$ condition).

Now we define the whole sequence by first choosing $N_j$'s that increase very rapidly, and then defining $c_k$ as above on each of these intervals, and $c_k=0 $ otherwise. Notice that for $t=1/N_j$ and $k$ taken from one of the other intervals $I_m$, $m\not= j$, then $e^{-\lambda_kt}-e^{-kt}$ is extremely small, either because $tk$ is either very small or very large. So these intervals make negligible contributions to (1).

It follows that (1) does not go to zero as $t\to 0+$ along the sequence $t=1/N_j$, and as discussed earlier, this means that we have a counterexample.

Finally, an obvious modification also gives examples where $c\in\ell^p$ for any (or all) $p>1$.

This is not true. Let's take $\lambda_k=k$. The problem is that for a given small $t>0$, there is a long intermediate interval where $e^{-kt}$ is no longer close to $1$, but not yet small enough to make $\sum |e^{-kt}c_k|$ negligible, and we can build a sequence $c_k$ to order that exploits this.

Let's focus on $t=1/N$ and $N\le k\le N\log N$. On the first half of this interval, let's take $c_k=\delta N^{-1/2}$, with a small $\delta>0$ to be determined later, and $c_k=-\delta N^{-1/2}$ on the second half of this interval. Then $$ \sum_{k=N}^{N\log N} e^{-k/N}c_k = e^{-1}\delta N^{1/2} + O(\delta) , $$ which of course is a disaster since $\sum_{k=N}^{N\log N} c_k = 0$.

We now just take a very rapidly increasing sequence $N_j$ and $\delta_j=1/j$, say, and then define $c_k$ as above on the corresponding intervals and $c_k=0$ otherwise. Then for $t=1/N_j$, we have $\sum e^{-kt} c_k \simeq N_j^{1/2}/j\not\to 0=\sum c_k$: for $N_j\le k\le N_j\log N_j$, the discussion above applies, and for the other intervals, $e^{-k/N_j}$ is either very close to $1$ or very small, so these sums are extremely close to zero.

Finally, note that $c\in\ell^2$, as desired; in fact, an obvious modification gives us examples with $c\in\ell^p$ for any (or all) $p>1$.

The problem with the example below is that $\sum c_k$ doesn't converge, and anyway it would contradict Abel's theorem on power series.

This is not true. Let's take $\lambda_k=k$.

Let's focus on $t=1/N$ and $N\le k\le N\log N$. On the first half of this interval, let's take $c_k=\delta N^{-1/2}$, with a small $\delta>0$ to be determined later, and $c_k=-\delta N^{-1/2}$ on the second half of this interval. Then $$ \sum_{k=N}^{N\log N} e^{-k/N}c_k = e^{-1}\delta N^{1/2} + O(\delta) , $$ which of course is a disaster since $\sum_{k=N}^{N\log N} c_k = 0$.

We now just take a very rapidly increasing sequence $N_j$ and $\delta_j=1/j$, say, and then define $c_k$ as above on the corresponding intervals and $c_k=0$ otherwise. Then for $t=1/N_j$, we have $\sum e^{-kt} c_k \simeq N_j^{1/2}/j\not\to 0=\sum c_k$: for $N_j\le k\le N_j\log N_j$, the discussion above applies, and for the other intervals, $e^{-k/N_j}$ is either very close to $1$ or very small, so these sums are extremely close to zero.

Finally, note that $c\in\ell^2$, as desired; in fact, an obvious modification gives us examples with $c\in\ell^p$ for any (or all) $p>1$.