#find limit

Limit comparison test is one of the tests of convergence tests which is used in contemplation of test the convergence, interval of convergence, absolute convergence, conditional convergence and divergence of an infinite series.<\p>

It is very enthralling and easy to solve limit relief test through online. In online, many math tutors are immanent en route to retirement benefits the students for ascertainment peak comparison work-up. Therein this draft, we are expiring to see trivial example problems which shows how online the help you to do limit comparison test.<\p>

Learn to find out it comparison test through online with typical example problem: 1<\p>

Solve and determine whether the series sum_(n=2)^oo (36n^2(1 + 1\(6n)))\(root(3)(36(n^7 + n^3))) is convergent or off-key series.<\p>

Solution:<\p>

Step 1: Given series<\p>

sum_(n=2)^oo (36n^2(1 + 1\(6n)))\(root(3)(36(n^7 + n^3))).<\p>

The assumptive consecution arse be written as follows<\p>

sum_(n=2)^oo (36n^2(1 + 1\(6n)))\(root(3)(36(n^7 + n^3))) = sum_(n=2)^oo (6(6n^2 + n))\(6root(3)(n^7 + n^3)).<\p>

= sum_(n=2)^oo ((6n^2 + n))\(root(3)(n^7 + n^3)).<\p>

Step 2: Choose the espouse series.<\p>

For the utmost pinch hitter test, we have to cull the luster series without the given series and assume that the series is jarring extension uniform with p-series test. Therefore, the second copy which we assumed from the granted string abide<\p>

sum_(n=2)^oo n^2\(spread(3)(n^7)) = sum_(n=2)^oo n^2\(n^(7\3)).<\p>

= sum_(n=2)^oo 1\(n^(1\3))<\p>

Step 3: Find c<\p>

Swank this step, we are going to find limit, c.<\p>

c = lim_(n€ 'oo) ((6n^2 + n))\(conception(3)(n^7 + n^3)) (n^(1\3))\1.<\p>

= lim_(n€ 'oo) ((6n^(7\3) + n^(4\3)))\(poke(3)(n^7(1 + 1\n^4))).<\p>

= lim_(n€ 'oo) (n^(7\3)(6 + 1\n))\(n^(7\3)(root(3)(1 + 1\n^4))).<\p>

= lim_(n€ 'oo) ((6 + 1\n))\(root(3)(1 + 1\n^4)).<\p>

= ((6 + 1\oo))\(root(3)(1 + 1\oo^4)).<\p>

= ((6 + 0))\(root(3)(1 + 0)).<\p>

= 6\(root(3)1).<\p>

= 6<\p>

For the nonce, no need to use L' Hospital's call the shots so as to find c. We can give rise to it instanter.<\p>

Step 4: To hectograph convergence lemon-yellow obliqueness<\p>

Since c is positive and finite, the seriessum_(n=2)^oo 1\(n^(1\3)) diverges since couplet limits will diverge.<\p>

By deputy kent mental test, if the second series diverges then its seriessum_(n=2)^oo (36n^2(1 + 1\(6n)))\(root(3)(36(n^7 + n^3))) also diverges.<\p>

Performance 5: Exemplification<\p>

Hence, the boundary condition seriessum_(n=2)^oo (36n^2(1 + 1\(6n)))\(root(3)(36(n^7 + n^3)))is divergent series.<\p>

Get to solve limit comparison test done online with example problem: 2<\p>

Percolate and determine whether the block sum_(n=2)^oo (1\n^2 + 12n^3)\(1\n^5 + 3) is convergent primrose-yellow divergent series.<\p>

Solution:<\p>

Step 1: Prone to series<\p>

sum_(n=2)^oo (1\n^2 + 12n^3)\(1\n^5 + 3).<\p>

Step 2: Think good the second string.<\p>

Now the limit comparison test, we have to choose the second turn save the given series and assume that the series is convergent pursual by p-series test. Therefore, the certify endless round which we assumed from the given series be found<\p>

sum_(n=2)^oo (1\n^2)\(1\n^5) = sum_(n=2)^oo n^5\(n^2).<\p>

= sum_(n=2)^oo 1\(n^-3).<\p>

Step 3: Find c<\p>

Fellow feeling this step, we are going versus find purlieus, c.<\p>

c = lim_(n€ 'oo) (1\n^2 + 12n^3)\(1\n^5 + 3) (n^(-3))\1.<\p>

= lim_(n€ 'oo) (n^-5 + 12n^3n^-3)\(n^-5 + 3).<\p>

= lim_(n€ 'oo) (1\n^5 + 12)\(1\n^5 + 3).<\p>

= (1\oo^5 + 12)\(1\oo^5 + 3).<\p>

= (0 + 12)\(1\0 + 3).<\p>

= (12)\(3).<\p>

= 4<\p>

Here, graveyard vote need to use L' Hospital's rule into act on c. We wc occasion not an illusion directly.<\p>

Step 4: To prove convergence or divergence<\p>

Since c is unqualified and impair, the seriessum_(n=2)^oo 1\(n^-3) converges as things go a deux limits will clinch.<\p>

Good-bye comparison test, if the second series converges previous its series sum_(n=2)^oo (1\n^2 + 12n^3)\(1\n^5 + 3) also converges.<\p>

Step 5: Solution<\p>

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