How do I find the value of the convergent series?
\[\sum_{n=1}^{\infty} \frac{ 1 }{ n(n+6)}\]
i would try partial fractions
and see if the series telescopes
im not sure how to do that.. :(
hmm... that is the only way I know... you sure you guys haven't talked about telescoping series?
i don't think i've learned about that
what have you guys mentioned in class
the topic is divergence and integral test
but those can't use to find the value of the series
oh...
do you know at least how to write your fraction as a sum of partial fractions ?
ok, i have 20 more min to do my online hw so can I show you another problem that might be easier?
i guess... so i guess that means you haven't done partial fractions here? Like you might remember them using them to ingrate certain things...
for example \[\frac{1}{n(n+6)}=\frac{A}{n}+\frac{B}{n+6}\] and you would combine the fractions on the right to find A and B
Test each of the following series for convergence by the Integral Test. If the integral test cannot be applied to the series, enter NA.\[\sum_{n=1}^{\infty}\frac{ n+5 }{ (-4)^n}\] ∑n=1∞n+5(−4)n
The integral test says: Suppose that f(n) is continuous, positive, and decreasing function on the interval [1,inf) then \[\int\limits_k^ \infty f(x) dx \text{ converges } \implies \sum_{n=k}^{\infty} f(n) \text{ converges } \\ \int\limits_k^\infty f(x) dx \text{ diverges } \implies \sum_{n=k}^{\infty} f(n) \text{ diverges }\] do you think that we satisfy the hypothesis part of this test ?
\[f(x)=\frac{x+5}{(-4)^{x}} \\ \text{ is this continuous on } [1,\infty) ? \\ \text{ is this positive on } [1,\infty)? \\ \text{ does this decrease on } [1,\infty)?\]
its decreasing, continuous, but not positive?
increasing
it is also not continuous for example at x=1/2 the denominator is not defined over real numbers but anyways we just need it to fail at least one of those 3 things to rule out the integral test and you definitely got that it wasn't completely positive on [1,inf)
ahh ok.
that's it?
it is your question :p It says use the integral test if it can be applied. If it cannot be applied then say NA.
was there suppose to be more to the question? because that is all I see...
ratio test might work here but we cannot use the integral test because it didn't meet the conditions to apply the integral test
or i mean alternating series test
ok thank you !
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