Now, while we have correctly identified this as a geometric series it doesn’t start at either of the two standard starting values of \(n\), i.e. \(n = 0\) or \(n = 1\).

This won’t stop us from determining if the series converges or diverges because that only depends on the value of \(r\) which we can determine regardless of the starting value of \(n\) with enough work. However, if the series does converge we won’t be able to use the formula for determining the value of the series as that also needs the value of \(a\) and that does require the series to start at one of the two standard starting values.

We have two options for taking care of this problem. One is to use an index shift to convert this into a series that starts at one of the standard starting values of \(n\). In most cases this is probably the only real option.

However, in this case let’s notice that this series is almost identical to the series from the previous problem. The only difference is that this series starts at \(n = 2\) while the series in the previous problem starts at \(n = 1\). This means that we can use the results of the previous problem to greatly reduce the amount of work needed here.

Show Step 3

We know that the series in the previous problem converged and since we’re only changing the starting value of \(n\) that will not affect the convergence of the series. Therefore, the series in this problem will also converge.

Since we also know that the value of the series in the previous series is \(\frac{{1225}}{2}\) we can find the value of the series in this problem. All we need to do is strip out one term from the series in the previous problem to get,

\[\sum\limits_{n = 1}^\infty {\frac{{{5^{n + 1}}}}{{{7^{n - 2}}}}} = \frac{{{5^2}}}{{{7^{ - 1}}}} + \sum\limits_{n = 2}^\infty {\frac{{{5^{n + 1}}}}{{{7^{n - 2}}}}} \]

Then using the value we found in the previous problem can get the value of the series from this problem as follows,

\[\frac{{1225}}{2} = 175 + \sum\limits_{n = 2}^\infty {\frac{{{5^{n + 1}}}}{{{7^{n - 2}}}}} \hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}\sum\limits_{n = 2}^\infty {\frac{{{5^{n + 1}}}}{{{7^{n - 2}}}}} = \frac{{1225}}{2} - 175 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{875}}{2}}}\]

On a quick side note if you did chose to do an index shift here are the two series (for each possible starting value of \(n\)) that you should have gotten.

\[\sum\limits_{n = 2}^\infty {\frac{{{5^{n + 1}}}}{{{7^{n - 2}}}}} = \sum\limits_{n = 1}^\infty {\frac{{{5^{n + 2}}}}{{{7^{n - 1}}}}} = \sum\limits_{n = 0}^\infty {\frac{{{5^{n + 3}}}}{{{7^n}}}} \]

Both of the last two are in the “standard” form and can be used to arrive at the same result as above.