Question

Solve 42-48 using ''the sandwich theorem'' @JamesJ @satellite73 and others please

Answer 1

i get -6

Answer 2

but i am also going to bet this is not the question

Answer 3

I want to discuss those examples 42-48 if you have time...and if you can see the examples :D

Answer 4

42, I would start like this...

Answer 5

1/sqrt(n^2+1)>1/sqrt(n^2+2)>.....>1/sqrt(n^2+2n)

Answer 6

Hence the sum is bounded by
\[ \frac{n}{\sqrt{n^2+1}} > S > \frac{n}{\sqrt{n^2+n}} \] and the limit of both of those is the same.

Answer 7

great, that's what I was going to ask, how do you know that the right term in the denominator is exactly n?

Answer 8

that's a little bit confusing for me in next examples? 1 2 3 and then it finishes with 2n?

Answer 9

ok being a bonehead i cannot deal with infinity first, so to see what is going on i used n = 5 and then maybe it will be clear. you get
\[\frac{1}{\sqrt{26}}+\frac{1}{\sqrt{27}}+...+\frac{1}{\sqrt{35}}\] now i can see how to get an upper bound and lower bound

Answer 10

OK

Answer 11

in this case i see that the upper bound is
\[\frac{1}{\sqrt{26}}+ ...+\frac{1}{\sqrt{26}}=\frac{5}{\sqrt{26}}\] and similarly the lower bound will be
\[\frac{5}{\sqrt{35}}\]

Answer 12

how did you get that 5?

Answer 13

i took five terms all to be the largest

Answer 14

because in the nth expression there are n terms in that expression.

Answer 15

lol except i got largest and smallest backward !!

Answer 16

sorry, couldn't log in

Answer 17

me too, some conection problems

Answer 18

ok well i hope this clears up question 42
jamesj wrote is succinctly

Answer 19

how about example 45?

Answer 20

In other words, consider the nth expression, call it S_n
1/sqrt(n^2 + 1) + 1/sqrt(n^2 + 2) + ...... + 1/sqrt(n^2 + n)

there are n terms here. ... oh, ok, you've got it?

Answer 21

yes, I see 1 2 3 ....n so there are n terms

Answer 22

but how would I know the number of terms in ex 45: sorry if I'm boring...:D

Answer 23

#45 looks ill defined to me.

Answer 24

In other words, it is not clear what the terms are.

Answer 25

i guess you keep adding until you get to
\[11n^2\]?

Answer 26

hmmm that's the thing that confuses me a lot...

Answer 27

in other examples too, I don't get how can I count the terms...? :(

Answer 28

oh ok. There's 11n^2 of them

Answer 29

not very clear, but i guess it is
1, 2, 3 , ... , n^2
n^2+1, ..., 2n^2
...
10n^2+1,..., 11n^2 making a total of
\[11n^2\] terms

Answer 30

can you somehow explain that to me James, how come 11n^2 when the number are 1,2,3...

Answer 31

you start with 1, you end with 11n^2
so there are 11n^2 terms

Answer 32

what sat73 said. Try it for n = 2 and you'll see the 22 terms

Answer 33

try it with n = 3

Answer 34

if you can't sleep tonight, try it for n = 17

Answer 35

tonight, tomorrow night, ...

Answer 36

haha :D can you be so kind and tell me the number of terms for all other examples...I 'll buy you something :D

Answer 37

does that mean that you will multiply with 11n^2?

Answer 38

for example for number 43, you start with 2 and end with 6n
try it with n = 5 and see how many terms you get
2,4,6,8,10,12,...,30
you will see there are 15 terms

Answer 39

where did the 15 come from?
\[\frac{6\times 5}{2}\]

Answer 40

I think I get it...

Answer 41

continue please with the rest...:D

Answer 42

so in general were will be for number 43, 3n terms

Answer 43

so I make the lower and the upper bound, and then multiply with the number of terms, that's how I'll make a ''sendwich'' I assume? :D

Answer 44

want to try 44?

Answer 45

yes, that is the idea for all of these i think

Answer 46

for 44, if I take n=3 I get 16 for n=4 I get 25 for n=5 I get 36...so the number of terms is actually n^2+2n+1?

Answer 47

hmm that is not what i got
take for example n =5

Answer 48

start with 25,26, 27 , ... , 36 so there are 36 - 25 = 11 terms

Answer 49

you are not starting at 1, you are starting an n^2

Answer 50

OMG I see that...lol :D

Answer 51

for that one since you are starting at n^2 and ending at (n+1)^2 i think there are
\[(n+1)^2-n^2=2n+1\] terms

Answer 52

the last one please and I'm done I promise :D