Vtyjkyuk

Study the pointwise, uniform, normal and on $L^p$ convergence of the series

$$sum_n=1^inftyfrace^-nx1+n^2$$

My approach

Let $f_n(x)=frace^-nx1+n^2$, it's easy to show that:

1. $f_n(x)ge 0 forall xinmathbbR$;




2. $lim_nto+inftyf_n(x)=0 iff xin [0,+infty)$




3. $sum_n=1^+inftyfrace^-nx1+n^2lesum_n=1^+inftyfrac1n^2+1lesum_n=1^+inftyfrac1n^2<+infty$

where the first inequality of (3) is true for $xge0$. From (3) I know that the series is normally convergent on $[0,+infty)$, hence it converges pointwise and uniformly on the $[0,+infty)$.

Now $sumfrace^-nxn^2+1$ is a normal convergent series in $[0,+infty)$ because $sumfrac1n^2$ converges.

Let $S_k(x)=sum_n=1^kfrace^-nx1+n^2$ and $S(x)=sum_n=1^+inftyfrace^-nx1+n^2$, how can I prove that

$$|S-S_k|_p^p=int_0^+inftyleft|sum_n=k+1^+inftyfrace^-nxn^2+1right|^pdx longrightarrow 0$$ for $kto +infty$.

My first idea is to use the Lebesgue dominated convergence theorem: if so, the limit operator goes trough the integral, hence $|S-S_k|_p^pto 0$ for $kto +infty$ because $$sum_n=k+1^+inftyfrace^-nxn^2+1$$ is the tail of a convergent series (then $sum_n=k+1^+inftyfrace^-nxn^2+1to 0$ for $kto+infty$).

Now I have many troubles to find $g(x)$ such that $left|sum_n=k+1^+inftyfrace^-nxn^2+1right|^ple g(x)$ and $g(x)in L^1([0,+infty))$

I try this:

$$sum_n=k+1^+inftyfrace^-nxn^2+1lesum_n=k+1^+inftye^-n x=frace^-(k+1)x1-e^-x$$

and now I'm stuck because of $1-e^-x$.

Every other inequalities I found are useless to find a proper $g(x)$. Maybe there is a easier way to solve this. Need help. Thanks.

Edit: This part

where the first inequality of (3) is true for $xge0$. From (3) I know
that the series is normally convergent on $[0,+infty)$, hence it
converges pointwise and uniformly on the $[0,+infty)$

is logically incorrect, in fact 3. must be $$3. frace^-nxn^2+1lefrac1n^2+1lefrac1n^2$$ and so $sumfrace^-nxn^2+1$ is a normal convergent series in $[0,+infty)$ because $sumfrac1n^2$ converges.

$$left|sum_n=k+1^+inftyfrace^-nxn^2+1right|^ple e^-px cdot left(sum_ngeq 1frac1(n^2+1)right)^p = c^p cdot e^-px$$
And $$ int_0^+infty c^p cdot e^-px = fracc^pp < + infty $$
So you can use dominated convergence theorem and show the $L^p$ convergence.