Determine whether the series is convergent or divergent
series symbol n=1 to infinity (n^2/(e^(3n))
an+1/an = e^-3 (1 + 2/n + 1/n^2)
that ratio is less than 1, so the series converges.
see
http://www.wolframalpha.com/input/?i=sum+n^2%2Fe^%283n%29
we can't use the ratio test on it. The only thing we know as of now is the integral test.
To determine whether the series is convergent or divergent, we can use the ratio test. The ratio test states that if the absolute value of the ratio of consecutive terms approaches a limit less than 1, then the series is convergent; if the ratio approaches a limit greater than 1 or infinity, then the series is divergent.
Let's apply the ratio test to the given series:
\[ \lim_{{n \to \infty}} \left| \frac{{a_{n+1}}}{{a_n}} \right| \]
Where \( a_n = \frac{{n^2}}{{e^{3n}}} \).
First, let's calculate \( a_{n+1} \):
\[ a_{n+1} = \frac{{(n+1)^2}}{{e^{3(n+1)}}} \]
Next, we find the ratio:
\[ \frac{{a_{n+1}}}{{a_n}} = \frac{{\frac{{(n+1)^2}}{{e^{3(n+1)}}}}}{{\frac{{n^2}}{{e^{3n}}}}} \]
Simplifying this expression, we get:
\[ \frac{{(n+1)^2}}{{n^2}} \cdot \frac{{e^{3n}}}{{e^{3(n+1)}}} \]
Further simplifying, we have:
\[ \frac{{n^2 + 2n + 1}}{{n^2}} \cdot \frac{{e^{3n}}}{{e^{3n+3}}} \]
Now, we can cancel out the common term:
\[ \frac{{1 + \frac{{2n}}{{n^2}} + \frac{{1}}{{n^2}}}}{1} \cdot \frac{{e^{3n}}}{{e^{3n+3}}} \]
Taking the limit as \( n \) approaches infinity:
\[ \lim_{{n \to \infty}} \left(\frac{{1 + \frac{{2n}}{{n^2}} + \frac{{1}}{{n^2}}}}{1} \cdot \frac{{e^{3n}}}{{e^{3n+3}}}\right) \]
Simplifying this limit:
\[ \lim_{{n \to \infty}} \left(\frac{{1 + \frac{{2}}{{n}} + \frac{{1}}{{n^2}}}}{1} \cdot \frac{{e^{3n}}}{{e^{3n} \cdot e^3}}\right) \]
Further simplification:
\[ \frac{{1 + 0 + 0}}{1} \cdot \frac{{1}}{{e^3}} \]
Which gives us:
\[ \frac{{1}}{{e^3}} \]
Since this ratio is less than 1, the series converges.
Therefore, the series \( \sum_{{n=1}}^{{\infty}} \frac{{n^2}}{{e^{3n}}} \) is convergent.