Question

show that the sequence is convergent

Answer 1

\[\huge{a_n=\sum_{k=1}^{n}\frac{ 1 }{ k+n }}\]

Answer 2

show that \[(a_n)^{\infty}_{n=1}\]
is convergent

Answer 3

look at each term in the sum, it's strictly bounded by 1/n. so the sum will be bounded by n*1/n = 1.

Answer 4

\[\frac{ 1 }{ n+1 }+\frac{ 1 }{ n+2 }+...+\frac{ 1 }{ 2n }\]

Answer 5

\[a_{n+1}<a_n\]

Answer 6

monotone decreasing

Answer 7

you dont need to show that... all you need to show is:
\[\lim_{n \rightarrow \infty} a_n < \infty\]

Answer 8

so\[a_n\] is bounded by 1/n,how I thought its 1/2n

Answer 9

\[\sum_{k = 1}^{n} \frac{ 1 }{ 2n } < \sum_{k = 1}^{n} \frac{ 1 }{ k+n } < \sum_{k = 1}^{n} \frac{ 1 }{ n }\]

Answer 10

lower bound doesnt really matter here (you can tell from a glance that a lower bound is 0). u just need the upper bound to be finite.

Answer 11

thanks making sense ,so is this done

Answer 12

well the LHS of the inequality is 1/2, and the RHS of the inequality is 1, so 1/2<a_n<1.
you can do the rest

Answer 13

awesome thanks,when dob

Answer 14

when do I check the ratio or differrence

Answer 15

i wonder if integration would be useful, just thinking maybe:
\[\lim_{n\to \infty}\int_{1}^{n}\frac{1}{n+x}dx=\lim_{n\to \infty}ln(\frac{2n}{1+n})=ln(2)\]
maybe?

Answer 16

since the continuous function converges, and it is bigger than the discrete function; the smaller discrete function converges as well since it is consumed within the bigger function

Answer 17

I think the continuous function you have is actually less than the discrete function (but i'm pretty sure you can shift it so that it becomes an upper bound)

Answer 18

hmm

\[\sum_{1}^{100}\frac{1}{k+100}=0.69\]
\[\int_{1}^{100}\frac{1}{x+100}=0.68\]

lol, thats odd

Answer 19

plot your continuous function and your discrete function and you'll see where the difference arises.