
MathHello! "Express" something in terms of `X` and `Y` means to find an expression (formula, rule) that gives this "something" as a result of operations on `X` and `Y.` The example of a formula is...

MathThe Limit Comparison Test states that if for two series with positive terms `sum a_n` and `sum b_n` the ratio `a_n/b_n` has a finite nonzero limit, then both series converge or both diverge. This...

MathThe Direct Comparison Test states that if for two series with nonnegative terms `sum a_n` and `sum b_n` the condition `a_n lt= b_n` holds for all `n` and `sum b_n` converges, then `sum a_n` also...

MathIf a series `sum a_n` converges, its term `a_n` tends to zero (the converse is not true in general). Thus if `a_n` does not tend to zero, the series is divergent. For this series `a_n = (1 +...

MathThe given series `sum_(n=0)^oo (2/3)^n` is in a form of geometric series. Recall that sum of geometric series follows the formula: `sum_(n=1)^oo a*r^(n1)` . or with an index shift: `sum_(n=0)^oo...

MathThe given series: `sum_(n=1)^oo 1/n^0.95` is in a form of pseries.` ` The test for pserires `sum_(n=1)^oo 1/n^p` follows the test for convergence as: if p is within the interval of `0ltplt=1`...

MathThis series is the sum of an infinite geometrical progression with the common ratio of `3/p.` It is well known that such a series converges if and only if its common ratio is less than `1` by the...

MathRecall that infinite series converges to single finite value `S` if the limit if the partial sum `S_n` as n approaches `oo` converges to `S` . We follow it in a formula: `lim_(ngtoo)...

MathRecall that infinite series converges to single finite value `S ` if the limit if the partial sum `S_n` as n approaches `oo` converges to `S` . We follow it in a formula: `lim_(ngtoo)...

Math`sum_(n=1)^oo1/(9n^2+3n2)` Let's rewrite the n'th term of the sequence as, `a_n=1/(9n^2+3n2)` `=1/(9n^2+6n3n2)` `=1/(3n(3n+2)1(3n+2))` `=1/((3n+2)(3n1))` Now let's carry out partial fraction...

Math`sum_(n=1)^oo(sin1)^n` Let's find the value of `sin1` from calculator. `sin(1)=0.84147<1` The series is geometric with first term (a) as `sin(1)` and common ratio (r) as `sin(1)` Geometric...

Math`sum_(n=0)^oo((0.3)^n+(0.8)^n)` `=sum_(n=0)^oo(0.3)^n+sum_(n=0)^oo(0.8)^n` Now both of the above are geometric series having first term as 1 and common ratios `0.3,0.8` respectively. Geometric...

Math`sum_(n=0)^oo(1/2^n1/3^n)` `=sum_(n=0)^oo1/2^nsum_(n=0)^oo1/3^n` `=(1/2^0+1/2^1+1/2^2+.......+1/2^oo)(1/3^0+1/3^1+1/3^2+......+1/3^oo)`...

Math`93+11/3+........` Let's check the common ratio of the terms, `r=a_2/a_1=3/9=1/3` `r=a_3/a_2=1/3=1/3` So this is a geometric sequence with common ratio `1/3` `S_oo=a_1/(1r)`...

Math`8+6+9/2+27/8+........` Let's find the common ratio of the terms: `r=a_2/a_1=6/8=3/4` `r=a_3/a_2=(9/2)/6=9/12=3/4` So this is a geometric sequence with common ratio of `3/4` `S_oo=a/(1r)` where a...

Math`sum_(n=1)^oo1/((2n+1)(2n+3))` Using partial fractions, we can write the n'th term of the sequence as, `a_n=1/(2(2n+1))1/(2(2n+3))` Now the n'th partial sum is,...

Math`sum_(n=1)^oo4/(n(n+2))` Using partial fractions we can write the n'th term of the series as, `a_n=4(1/(2n)1/(2(n+2)))` `a_n=(2/n2/(n+2))` Now the n'th partial sum is,...

MathBy the definition, the sum of a series is the limit of its partial sums (if it exists). For this series the `N`th partial sum is `sum_(n=0)^N (1/5)^n.` This sum is the sum of finite geometric...

Math`sum_(n=0)^oo5(2/3)^n` `=5((2/3)^0+(2/3)^1+(2/3)^2+.........+(2/3)^oo)` `=5(1+(2/3)^1+(2/3)^2+........(2/3)^oo)` This is a geometric series with first term (a) as `1` and common ratio (r) as...

Math`sum_(n=0)^oo (0.6)^n` To verify if the series is convergent, apply the Geometric Series Test. Take note that a geometric series has a form `sum` `ar^n` It converges if `rlt1` . And it diverges...

Math`sum_(n=0)^oo (0.9)^n` To verify if the series is convergent, apply the Geometric Series Test. Take note that a geometric series has a form `sum` `ar^n` It converges if `rlt1` . And it...

MathTo verify if the given infinite series: `sum_(n=1)^oo 2(1/2)^n` converges, recall that infinite series converges to single finite value S if the limit if the partial sum `S_n ` as n approaches`...

Math`sum_(n=0)^oo (5/6)^n` To verify if the series is convergent, apply the Geometric Series Test. Take note that a geometric series has a form `sum_(n=0)^oo ar^n` or `sum_(n=1)^oo...

Math`sum_(n=1)^oo (n!)/2^n` To verify if the series diverges, apply the ratio test. The formula for ratio test is: `L = lim_(n>oo) a_(n+1)/a_n` If L<1, the series converges. If L>1, the...

Math`sum_(n=1)^oo n/sqrt(n^2+1)` To verify if the series diverges, apply the nthTerm Test for Divergence. It states that if the limit of {an} is not zero, or does not exist, then the sum diverges....

Math`sum_(n=1)^oo n^2/(n^2+1)` To verify if the series diverges, apply the nthTerm Test for Divergence. It states that if the limit of {an} is not zero, or does not exist, then the sum diverges....

Math`sum_(n=1)^oo n/(2n+3)` To verify if the series diverges, apply the nthTerm Test for Divergence. It states that if the sequence {an} does not converge to zero, then the series diverges....

Math`sum_(n=1)^oo n/(n+1)` To verify if this infinite series diverges, apply the Divergent Test. In Divergence Test, it states that if the limit of an is not zero, or does not exist, then the sum...

MathRecall that infinite series converges to single finite value `S` if the limit if the partial sum `S_n ` as n approaches `oo` converges to `S` . We follow it in a formula: `lim_(ngtoo)...

Math`sum_(n=0) (7/6)^n` To verify if the series is divergent, apply the Geometric Series Test. Take note that a geometric series has a form `sum ar^n` It converges if `rlt1` . And it diverges if...

Math`a_n=(3/2)^n` A sequence is monotone if it is increasing for all n, or decreasing for all n. `a_(n+1)=(3/2)^(n+1)` `a_(n+1)=(3/2)^n(3/2)` `a_(n+1)=(3/2)a_n` So, `a_(n+1)>a_n` So the sequence is...

MathA sequence is monotone if it is increasing for all n , or decreasing for all n. We are given:`a_n=(2/3)^n` `a_(n+1)=(2/3)^(n+1)` `a_(n+1)=(2/3)^n(2/3)` `a_(n+1)=a_n(2/3)` So, `a_(n+1)<a_n` So...

MathA sequence is monotone if it is increasing for all n , or decreasing for all n. We are given:`a_n=41/n` `a_(n+1)=41/(n+1)` `a_n<a_(n+1)` for all n. `a_n<4` So the sequence is monotonic...

Math`1/(2*3),2/(3*4),3/(4*5),4/(5*6)` From the numbers we can see that the numerator starts from 1 for the first number and increases by step of 1. Denominator follows the pattern: `(n+1)(n+2)` We can...

Math`2,1+1/2,1+1/3,1+1/4,1+1/5` The first term can be written as `1+1/1` So from the pattern of the numbers, n'th term of the sequence can be written as: `a_n=1+1/n` `a_n=(n+1)/n`

MathGiven the sequence 2/3, 3/4, 4/5, 5/6, ... An expression for the nth term of the sequence is `a_n=(n+1)/(n+2)` when n starts with 1. `a_1=(1+1)/(1+2)=2/3` `a_2=(2+1)/(2+2)=3/4`...

Math`1,1/4,1/9,1/16` The sequence can be written as, `1/1^2,1(1/2^2),1/3^2,1(1/4^2)` Based on the above pattern, we can write the n'th term of the sequence as, `a_n=(1)^(n1)(1/n^2)`

Math`2,1,6,13,22,...........` The terms of the sequence can be written as , `(1^23),(2^23),(3^23),(4^23),(5^23),..........` So from the above pattern of the terms we can write down the n'th term...

MathConsider the ratios between adjacent terms of the given sequence: `1/2 : 1 = 1/2,` `1/6 : 1/2 = 1/3,` `1/24 : 1/6 = 1/4,` `1/120 : 1/24 = 1/5.` We see that the `n` th term of the sequence may be...

MathGiven the sequence: 2, 8, 14, 20, ... The given sequence is an arithmetic sequence. The sequence is arithmetic because the common difference between each term is 6. In this sequence the common...

Math`a_n=cos(2/n)` To determine the limit of this function, let the n approaches infinity. `lim_(n>oo) a_n` `=lim_(n> oo) cos(2/n)` To solve, let the angle 2/n be equal to u, `u = 2/n` . Take...

Math`a_n=(2n)/sqrt(n^2+1)` To find the limit of a sequence, let the n approaches infinity. `lim_(n>oo) a_n` `=lim_(n>oo) (2n)/sqrt(n^2+1)` `= lim_(n>oo) (2n)/sqrt(n^2(1+1/n^2))`...

Math`a_n=6+2/n^2` To find the limit of a sequence, let the n approaches infinity. `lim_(n>oo) a_n` `=lim_(n>oo) (6 + 2/n^2)` `=lim_(n>oo) 6 + lim_(n>oo) 2/n^2` Take note that a limit...

Math`a_n = (5n^2)/(n^2+2)` To determine the limit of this sequence, let the n approaches infinity. `lim_(n>oo) a_n` `=lim _(n>oo) (5n^2)/(n^2+2)` To solve, factor out the n^2 in the...

Math`6, 2, 2/3, 2/9` To determine the next two terms, identify if it is an arithmetic or geometric sequence. Take note that an arithmetic sequence has a common difference. While a geometric sequence...

MathThe given sequence is 5, 10, 20, 40, ... Final Answer: The next two apparent terms are 80 and 160. The pattern to find the next term is to multiply the previous term by 2. The common ratio between...

MathThe given sequence is 8, 13, 18, 23, 28... Final Answer: The next two apparent terms are 33 and 38. The pattern to find the next term in the sequence is to add 5 to the previous term. The common...

MathGiven the sequence 2, 5, 8, 11, ... Final answer: The next two apparent terms are 14 and 17. The pattern to find the next term is to add 3 to the previous term. The common difference between each...

Math`a_1=6,a_(k+1)=1/3a_k^2` `a_2=1/3a_1^2=1/3(6)^2=1/3(36)=12` `a_3=1/3a_2^2=1/3(12)^2=144/3=48` `a_4=1/3a_3^2=1/3(48)^2=1/3(2304)=768` `a_5=1/3a_4^2=1/3(768)^2=1/3(589824)=196608` The first five...

Math`a_1=1` ,`a_(k+1)=2(a_k1)` `a_(1+1)=2(a_11)` `a_2=2(11)=2(0)=0` `a_3=2(a_21)=2(01)=2(1)=2` `a_4=2(a_31)=2(21)=2(3)=6` `a_5=2(a_41)=2(61)=2(7)=14` The first five terms of the...
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