does the sum of [(-5)^n]/[13^(n+1)] converge? as n goes to infinity
I think that since it oscillates it does not go to infinity. and therefore the series diverges.
To determine if the sum of the sequence \(\left(\frac{{(-5)^n}}{{13^{n+1}}}\right)\) converges or diverges, we can apply the ratio test.
The ratio test states that if for a series \(\sum a_n\), the limit \(\lim_{{n \to \infty}} \left|\frac{{a_{n+1}}}{{a_n}}\right|\) exists and is less than 1, then the series converges. If the limit is greater than 1 or infinite, the series diverges.
Let's apply the ratio test to the given series.
First, find the general term of the sequence:
\(a_n = \frac{{(-5)^n}}{{13^{n+1}}}\)
Next, calculate the ratio of consecutive terms:
\(\frac{{a_{n+1}}}{{a_n}} = \frac{{\frac{{(-5)^{n+1}}}{{13^{(n+1)+1}}}}}{{\frac{{(-5)^n}}{{13^{n+1}}}}} = \frac{{(-5)^{n+1} \cdot 13^{n+1}}}{{(-5)^n \cdot 13^{n+1} \cdot 13}}\)
Notice that \(13^{n+1}\) cancels out.
\(\frac{{a_{n+1}}}{{a_n}} = \frac{{(-5)(-5)^n}}{{13}}\)
Now, take the absolute value of this expression:
\(\left|\frac{{a_{n+1}}}{{a_n}}\right| = \frac{{5 \cdot 5^n}}{{13}}\)
As \(n\) approaches infinity, the expression \(\left|\frac{{a_{n+1}}}{{a_n}}\right|\) becomes:
\(\lim_{{n \to \infty}} \left|\frac{{5 \cdot 5^n}}{{13}}\right|\)
Since \(5^n\) grows without bound as \(n\) approaches infinity, the ratio \(\left|\frac{{5 \cdot 5^n}}{{13}}\right|\) also grows without bound.
Therefore, the limit of \(\left|\frac{{a_{n+1}}}{{a_n}}\right|\) is greater than 1, indicating that the series diverges.
Hence, the sum of the series \(\left(\frac{{(-5)^n}}{{13^{n+1}}}\right)\) diverges as \(n\) goes to infinity.