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Question

(3 Use the Limit Comparison Test to determine whether the series converges. points each) 3n 2 n 5 3n2 1 n 3

Solution

5 (1 Ratings )

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**Question**

(3 Use the Limit Comparison Test to determine whether the series converges. points each) 3n 2 n 5 3n2 1 n 3

** **

**Solution**

2)

a)_{[n=1 to ]} (n+4)_{/(3n+2)}

a_{n}=(n+4)_{/(3n+2)},b_{n}=1_{/n}

lim_{[n->]}an_{/bn}

=lim_{[n->]}[(n+4)/(3n+2)]_{/(1/n)}

=lim_{[n->]}[n(n+4)_{/(3n+2)}]

=lim_{[n->]}[n(1+(4/n))_{/n(3+(2/n))}]

=lim_{[n->]}[(1+(4/n))_{/(3+(2/n))}]

=[(1+0) _{/(3+0)}]

=1_{/3}

_{[n=1 to ]} 1_{/n} diverges by p series test , p=1_{/2}<1

so by limit comparision test _{[n=1 to ]} (n+4)_{/(3n+2)} diverges

---------------------------------

b)

_{[n=1 to ]} tan^{-1}(n)_{/(n+5)}

a_{n}=tan^{-1}(n)_{/(n+5)},b_{n}=1_{/n}

lim_{[n->]}an_{/bn}

=lim_{[n->]}[tan^{-1}(n)_{/(n+5)}]/(1_{/n})

=lim_{[n->]}[ntan^{-1}(n)_{/(n+5)}]

=lim_{[n->]}[ntan^{-1}(n)_{/n(1+(5/n))}]

=lim_{[n->]}[tan^{-1}(n)_{/(1+(5/n))}]

=[(/2)_{/(1+0)}]

=_{/2}

_{[n=1 to ]} 1_{/n} diverges by p series test , p=1

so by limit comparision test _{[n=1 to ]} tan^{-1}(n)_{/(n+5)} diverges

----------------------------------

c)_{[n=1 to ]} (5n^{4}-3n^{2}+1)_{/(n}^{6}+n^{3}+3)

a_{n}=(5n^{4}-3n^{2}+1)_{/(n}^{6}+n^{3}+3),b_{n}=1_{/n}^{2}

lim_{[n->]}an_{/bn}

=lim_{[n->]}[(5n^{4}-3n^{2}+1)_{/(n}^{6}+n^{3}+3)]_{/(1/n}^{2})

=lim_{[n->]}[n^{2}(5n^{4}-3n^{2}+1)_{/(n}^{6}+n^{3}+3)]

=lim_{[n->]}[n^{6}(5-(3/n^{2})+(1/n^{4}))_{/n}^{6}(1+(1/n^{3})+(3/n^{6}))]

=lim_{[n->]}[(5-(3/n^{2})+(1/n^{4}))_{/(1+(1/n}^{3})+(3/n^{6}))]

=[(5-0+0)_{/(1+0+0)}]

=5

_{[n=1 to ]} 1_{/n}^{2} converges by p series test , p=2>1

so by limit comparision test_{[n=1 to ]} (5n^{4}-3n^{2}+1)_{/(n}^{6}+n^{3}+3) converges

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