MATH SOLVE

4 months ago

Q:
# (1 point) by recognizing each series below as a taylor series evaluated at a particular value of x, find the sum of each convergent series.a. 1+2+222!+233!+244!+⋯+2nn!+⋯=b. 1−422!+444!−466!+⋯+(−1)n42n(2n)!+⋯=

Accepted Solution

A:

Guessing on what you're trying to say:

(a)

[tex]1+2+\dfrac{2^2}{2!}+\dfrac{2^3}{3!}+\cdots+\dfrac{2^n}{n!}+\cdots=\displaystyle\sum_{n=0}^\infty\frac{2^n}{n!}[/tex]

Compare to the series

[tex]e^x=\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!}[/tex]

which means the value of the sum above is [tex]e^2[/tex].

(b)

[tex]1-\dfrac{4^2}{2!}+\dfrac{4^4}{4!}-\cdots+\dfrac{(-1)^n4^{2n}}{(2n)!}+\cdots=\displaystyle\sum_{n=0}^\infty\frac{(-1)^n4^{2n}}{(2n)!}[/tex]

Compare to

[tex]\cos x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n)!}[/tex]

so that the sum evaluates to [tex]\cos4[/tex].

(a)

[tex]1+2+\dfrac{2^2}{2!}+\dfrac{2^3}{3!}+\cdots+\dfrac{2^n}{n!}+\cdots=\displaystyle\sum_{n=0}^\infty\frac{2^n}{n!}[/tex]

Compare to the series

[tex]e^x=\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!}[/tex]

which means the value of the sum above is [tex]e^2[/tex].

(b)

[tex]1-\dfrac{4^2}{2!}+\dfrac{4^4}{4!}-\cdots+\dfrac{(-1)^n4^{2n}}{(2n)!}+\cdots=\displaystyle\sum_{n=0}^\infty\frac{(-1)^n4^{2n}}{(2n)!}[/tex]

Compare to

[tex]\cos x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n)!}[/tex]

so that the sum evaluates to [tex]\cos4[/tex].