How do you use the limit comparison test to determine if #Sigma n/((n+1)2^(n-1))# from #[1,oo)# is convergent or divergent?

Question

Scarlett Allison · Accepted Answer

#sum_(n=1)^oon/((n+1)2^(n-1))# Rewriting this: #=sum_(n=1)^oon/(n+1)1/(2^(n-1))=sum_(n=1)^oon/(n+1)1/(2^n/2)=sum_(n=1)^oon/(n+1)2/2^n# Bringing the #2# out and noticing that #1/2^n=(1/2)^n#: #=2sum_(n=1)^oon/(n+1)(1/2)^n# We should recognize that #sum_(n=1)^oo(1/2)^n# is a geometric series, and since #1/2<1#, we know this series will converge. Also note that when #n>0#, all terms of #n/(n+1)<1#. This means we can say that: #n/(n+1)(1/2)^n<=(1/2)^n# We can now use the direct comparison test. Since #sum_(n=1)^oo(1/2)^n# converges, and #n/(n+1)(1/2)^n<=(1/2)^n#, we know that #sum_(n=1)^oon/(n+1)(1/2)^n# converges as well. Thus #2sum_(n=1)^oon/(n+1)(1/2)^n=sum_(n=1)^oon/((n+1)2^(n-1))# is convergent.

Elijah Abram · Answer

undefined To use the limit comparison test to determine if the series $ \sum_{n=1}^{\infty} \frac{n}{(n+1)2^{n-1}} $ is convergent or divergent:

1. Choose a series $ \sum_{n=1}^{\infty} b_n $ that is known to converge and is positive for all $ n $.

2. Calculate the limit:
$$ \lim_{n \to \infty} \frac{a_n}{b_n} $$
where $ a_n = \frac{n}{(n+1)2^{n-1}} $.

3. If the limit is a positive finite number, then both series either converge or diverge together. If the limit is zero or infinity, then the comparison is inconclusive.

4. Based on the comparison, if $ \sum_{n=1}^{\infty} b_n $ converges, then $ \sum_{n=1}^{\infty} a_n $ also converges. If $ \sum_{n=1}^{\infty} b_n $ diverges, then $ \sum_{n=1}^{\infty} a_n $ also diverges.

In this case, let's choose $ b_n = \frac{1}{2^n} $ since it's a geometric series that is known to converge.

Then, we calculate the limit:
$$ \lim_{n \to \infty} \frac{\frac{n}{(n+1)2^{n-1}}}{\frac{1}{2^n}} $$
$$ = \lim_{n \to \infty} \frac{2n}{(n+1)} $$

Now, simplify the limit:
$$ = \lim_{n \to \infty} \frac{2}{1 + \frac{1}{n}} $$
$$ = 2 $$

Since the limit is a positive finite number, and $ \sum_{n=1}^{\infty} \frac{1}{2^n} $ converges (it's a geometric series with $ r = \frac{1}{2} < 1 $), by the limit comparison test, $ \sum_{n=1}^{\infty} \frac{n}{(n+1)2^{n-1}} $ also converges.