Shifting trick in double series show that if $\displaystyle e^{x/2(t+1/t)}=\displaystyle\sum\limits_{n=-\infty}^\infty I_n t^n$ then $\displaystyle I_n=\sum\limits_{n=0}^\infty \frac{\left(x/2\right)^n}{(n+k)!k!}$

It was an exam question, lecturer never gives hint and answer even after the exam, so there might be little mistakes in the question.

My aim to ask this question is in the following direct solution somewhere they use shifting method go from n-finite to infinite sum limits. Any further and related hint, information will be appreciated.

$$\displaystyle e^{x/2(t+1/t)}=\displaystyle\sum\limits_{n=0}^{\infty}\left(x/2.\right(t+1/t))^n/n!$$$$=\displaystyle\sum\limits_{n=0}^{\infty}\left(\dfrac x2\right)^n/n!\sum_{k=0}^n\dbinom{n}{k}t^{n-k}t^{-k}$$$$=\displaystyle\sum\limits_{n=0}^{\infty}\sum_{k=0}^n\dfrac{(x/2)^n}{(n-k)!k!}t^{n-2k}$$

I stuck here, how can I get $\sum_0^\infty\sum_0^\infty$


Your formula for $I_n$ has the wrong power of $(x/2)$, and you should be summing over $k$ instead of $n$.

We have \begin{align*}\require{color} \sum_{n=-\infty}^\infty I_nt^n &=e^{x/2(t+1/t)}\\ &=\sum_{m=0}^\infty\sum_{k=0}^m\frac{(x/2)^m}{(m-k)!k!}t^{m-2k}\\ &=\sum_{k=0}^\infty\sum_{m=k}^\infty\frac{(x/2)^m}{(m-k)!k!}t^{m-2k}&&(\text{interchange order})\\ &=\sum_{k=0}^\infty\sum_{n=-k}^\infty\frac{(x/2)^{n+2k}}{(n+k)!k!}t^{n}&&(\text{let }m-2k=n)\\ &=\sum_{n=-\infty}^\infty{\color{blue}\sum_{k=0}^{\infty}\frac{(x/2)^{n+2k}}{(n+k)!k!}}t^n&&(\text{interchange order})\\ \end{align*} Since $n+k<0\Rightarrow \frac1{(n+k)!}=\frac1{\Gamma(n+k+1)}=0$, there is no need to restrict $n+k\geq 0$ as would have expected in the last step. So equating coefficients gives $$ I_n=\sum_{k=0}^{\infty}\frac{(x/2)^{n+2k}}{(n+k)!k!} $$

Remark: It is a little easier if you started off by $$ \exp((x/2)(t+1/t))=\exp((x/2)t)\exp((x/2)/t) $$ and multiply the two exponential series. You can see explicitly the $(n+k)!$ and $k!$ come from matching powers of $t$ for every $n$, as well as the $(x/2)^{n+2k}$.


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