How does it hold that $\lim_{k\to+\infty}f_k(x)=1?$

Question

Let $f_k(x)= 1-\frac {x^2} k +\frac {x^4}{2!k(k+1)}-\frac {x^6}{3!k(k+1)(k+2)}+...$
,$\forall x\in \mathbb R, k\notin \{0,-1,-2,-3,...\}$

Claim: For each $x\in \mathbb R , \quad \lim_{k\to+\infty}f_k(x)=1.$

Proof:In fact, the sequence ${x^{2n}\over n!}$ is bounded (since it converges to $0$) and if C is an upper bound and if $k\gt 1$ then
$\bigl|f_k(x)-1\bigr|\leqslant\sum_{n=1}^\infty\frac C{k^n}=C\frac{1/k}{1-1/k}=\frac C{k-1}.$

In this proof, the convergence of the sequence ${x^{2n}\over n!} $ is used. But in $f_k(x)$ there is the sequence $(-1)^n{x^{2n}\over n!}$

And I know since $\sum_{n=0}^\infty {x^{2n}\over n!} =e^{x^2}$ i.e this series converges, the sequence ${x^{2n}\over n!}$ must tend to $0$. But the sequence in $f_k(x) $ is $(-1)^n {x^{2n}\over n!}$ so doesn't this change anything? How does this imply that $\lim_{k\to+\infty}f_k(x)=1?$ I mean intuitively I see that as $k\rightarrow \infty $ limit of $f_k(x)$ goes to 1, but how is it proved by using the fact that the sequence ${x^{2n}\over n!}$ converges to $0?$ Do I misinterpret something ? May someone explain the process of thinking made in the proof ?

Answer 1

Since $k(k+1)...(k+n) \geq k$ we get $|f_k (x)-1| \leq \frac 1 k \sum \frac {x^{2n}} {n!}$. The last sum is convergent and we are done.

Answer 2

Since $$ \left|\,\frac{(-1)^j(k-1)!}{j!(j+k-1)!}x^{2j}\,\right|\le\frac{x^{2j}}{j!}\tag1 $$ and $$ \sum_{j=0}^\infty\frac{x^{2j}}{j!}=e^{x^2}\tag2 $$ converges for all $x$, our series is dominated by a convergent series. Furthermore, for all $x$ and each $j\ge1$, $$ \lim_{k\to\infty}\frac{(-1)^j(k-1)!}{j!(j+k-1)!}x^{2j}=0\tag3 $$ and for all $x$ and $j=0$, $$ \frac{(-1)^j(k-1)!}{j!(j+k-1)!}x^{2j}=1\tag4 $$ Thus, by Dominated Convergence, we can commute the limit with the sum: $$ \begin{align} \lim_{k\to\infty}\sum_{j=0}^\infty\frac{(-1)^j(k-1)!}{j!(j+k-1)!}x^{2j} &=\sum_{j=0}^\infty\lim_{k\to\infty}\frac{(-1)^j(k-1)!}{j!(j+k-1)!}x^{2j}\\ &=\overbrace{\ \ \ \ 1\ \ \ \ \vphantom{\sum_{j=0}^\infty}}^{j=0\text{ term}}+\sum_{j=1}^\infty0\\[9pt] &=1 \tag5 \end{align} $$

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Dominated Convergence for Series

Suppose that we have a sequence of series $$ s_n=\sum_{k=1}^\infty a_{n,k}\tag6 $$ where, for each $k$, $$ a_k=\lim_{n\to\infty}a_{n,k}\tag7 $$ exists and, for all $n$, $|a_{n,k}|\le b_k$ where $$ \sum_{k=1}^\infty b_k\lt\infty\tag8 $$ Then the limit and the sum commute; that is, $$ \lim_{n\to\infty} s_n=\sum_{k=1}^\infty a_k\tag9 $$ Proof: Note that $|a_k|\le b_k$. Choose $\epsilon\gt0$.

By $(8)$, we can find a $K$ large enough so that $$ \sum_{k\gt K}b_k\le\frac\epsilon4 $$ therefore, for all $n$ $$ \begin{align} \left|\,\sum_{k\gt K}(a_{n,k}-a_k)\,\right| &\le\sum_{k\gt K}|a_{n,k}-a_k|\\ &\le\frac\epsilon2 \end{align} $$ Now, by $(7)$, we can find an $N$ so that for $k\le K$ and $n\ge N$, $$ |a_{n,k}-a_k|\le\frac{\epsilon}{2K} $$ Therefore, $$ \begin{align} \left|\,\sum_{k=1}^\infty(a_{n,k}-a_k)\,\right| &\le\sum_{k=1}^\infty|a_{n,k}-a_k|\\ &\le K\frac{\epsilon}{2K}+\frac\epsilon2\\[6pt] &=\epsilon \end{align} $$

Answer 3

We have $f_k(x)= 1-\frac {x^2} k +\frac {x^4}{2!k(k+1)}-\frac {x^6}{3!k(k+1)(k+2)}+...$. let's set $C_j=\frac{x^{2j}}{j!}$.

So we have $f_k(x)=1-C_1\frac1k+C_2\frac1{k(k+1)}-C_3\frac1{k(k+1)(k+2)}...$

Now we need to notice that $$\frac1{k(k+1)(k+2)\cdots(k+j-1)}=\frac{(k-1)!}{(k-1)!\times(k+1)(k+2)\cdots(k+j-1)}=\frac{(k-1)!}{(k-1+j)!}$$let's set $K_j=\frac{(k-1)!}{(k-1+j)!}$. So we get $$f_k(x)=\overbrace{\overbrace{K_0}^1\overbrace{C_0}^1}^1-K_1C_1+K_2C_2...$$

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Now we let's use the fact that $\sum_{j=0}^\infty C_j=\sum_{j=0}^\infty \frac{x^{2j}}{j!}=e^{x^2}$. We know that $C_j\ge0$ for all $j$, and that the sum converges, hence $C_j$ exists, we don't care about the value, only the fact it is a real number.

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Now let's look on $K_j$:

$j>0$:

We know that $0<K_{j+1}\le K_{j}$ for all $j$(can you tell why?), So let's look on $\lim_{k\to\infty}K_1=\lim_{k\to\infty}\frac{(k-1)!}{(k-1+1)!}=\lim_{k\to\infty}\frac{(k-1)!}{(k)!}=\lim_{k\to\infty}\frac{1}{k}=0$. Because $K_{j+1}\le K_{j}$ we call conclude that $\lim_{k\to\infty}K_j=0$ for all $j>0$.

$j=0:$

$K_0=\frac{(k-1)!}{(k-1)!}=1$

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Let's go back to our series,

We have $f_k(x)=K_0C_0-K_1C_1+K_2C_2...=\sum_{j=0}^\infty (-1)^jK_jC_j=(-1)^0K_0C_0+\sum_{j=1}^\infty (-1)^jK_jC_j$

We already conclude that $C_j$ is a real number and that $\lim_{k\to\infty}K_j=0,j>0$, a real number times $0$ equal $0$, so we get $$\lim_{k\to\infty}f_k(x)=\lim_{k\to\infty}(-1)^0K_0C_0+\lim_{k\to\infty}\sum_{j=1}^\infty (-1)^jK_jC_j=\lim_{k\to\infty}(-1)^0K_0C_0+\lim_{k\to\infty}\sum_{j=1}^\infty (-1)^j0=\lim_{k\to\infty}(-1)^0K_0C_0+\sum_{j=1}^\infty 0=\lim_{k\to\infty}(-1)^0K_0C_0$$ we can calculate and see that $\lim_{k\to\infty}(C_0=1,K_0=1,(-1)^0=1)$ and conclude that $$\boxed{\lim_{k\to\infty}f_k(x)=1\times1\times1=1}$$