Title: Limits and series
Course: Demonstrating STACK (v4.3.7)

Find the limit, as \(n\) tends to plus infinity, of the following sequence, if a limit exists:

\[ a_{n} = [value] \sin\left(\frac{n \pi}{2}\right) . \]

If no limit exists, enter 11111.

limit =
Feedback: As \(n\) tends to plus infinity, the following sequence:\[a_{n} = [value] \sin\left(\frac{n \pi}{2}\right)\] cycles through 0, +[value], 0 and -[value] and never reaches a limit.

Find the limit, as \(n\) tends to plus infinity, of the following sequence, if a limit exists:

\[ a_{n} = [value] \cos\left(\frac{(2 n -1) \pi}{2}\right) . \]

If no limit exists, enter 11111.

limit =
Feedback: Find the limit, as \(n\) tends to plus infinity, of the following sequence, if a limit exists: \[ a_{n} = [value] \cos\left(\frac{(2 n -1) \pi}{2}\right) . \] \(a_{n} = 0\) for all \(n\), therefore the limit is equal to 0

Find

\[[value] + \frac{[value] \pi^2}{2!} - \frac{[value] \pi^4}{4!} +\frac{[value] \pi^6}{6!} - \ldots.\]

Answer:
Feedback: \[[value] + \frac{[value] \pi^2}{2!} - \frac{[value] \pi^4}{4!} +\frac{[value] \pi^6}{6!} - \ldots.\] can be written as \[[value](1 - \frac{[value] \pi^2}{2!} + \frac{[value] \pi^4}{4!} -\frac{[value] \pi^6}{6!} + \ldots.)\] \[[value](1 - \frac{([value] \pi)^2}{2!} + \frac{([value] \pi)^4}{4!} -\frac{([value] \pi)^6}{6!} + \ldots.)\] which equals \[[value] *\cos([value] \pi) = [value]\]

Using the Binomial series, find the first four non zero terms in the Maclaurin series for

 \[\sqrt{[value]+[value]x}\]

Answer:
Feedback: \[\sqrt{[value]+[value]x}\] can be written as \[[value]\sqrt{\left(1+[value]\right)}.\]As \[ {(1+x)}^{1/2} = 1 +\frac{1}{2} x + \frac{(1/2)(1/2-1)}{2 !} x^2 +\frac{(1/2)(1/2 -1)(1/2 - 2)}{3 ! } x^3 + \ldots ,\] \[ {(1+ [value] )}^{1/2} = 1 +\frac{1}{2} [value] + \frac{(1/2)(1/2-1)}{2 !} ([value] )^2 +\frac{(1/2)(1/2 -1)(1/2 - 2)}{3 ! } ([value] )^3 + \ldots .\] Hence \[\sqrt{[value]+[value]x} = [value](1 + [value] - [value] + [value] + \ldots) \] or\[\sqrt{[value]+[value]x} = [value] +[value] x-[value] x^2 + [value] x^3 + \ldots \]

By writing \(x\) as \([value]+(x-[value])\) and using the Maclaurin expansion for \(\ln(1+t)\), or otherwise, find the first four non zero terms in the Taylor series expansion for

\[\ln([value]x)\]

about \(x = [value]\).

Use exact expressions, i.e. enter \(\ln(5)\) rather than \(1.609...\).

Answer:
Feedback: \[\ln([value]x)\] can be written as \[\ln([value]([value]+(x-[value]))) \] which equals \[ \ln[value] + \ln([value]+(x-[value])) =\ln[value] + \ln\left([value]\left(1+\frac{(x-[value])}[value]\right)\right)=\ln[value] + \ln[value]+\ln\left(1+\frac{(x-[value])}[value]\right)\] As \(\ln(1+t) = t-\frac{t^2}{2}+\frac{t^3}{3}\) and \(\ln[value] + \ln[value] =\ln([value])\) \[\ln([value]x)= \ln([value])+\frac{x-[value]}[value]-\frac{(x-[value])^2}[value]+\frac{(x-[value])^3}[value]\]