mathematical induction

[erblina]

can somebody help me to prove this inequality using the mathematical induction :

n < [n/(2n-1)+(n+1)/(2n+1)+(n+2)/(2n+3)+⋯+(3n-3)/(6n-7)+(3n-2)/(6n-5)+(3n-1)/(6n-3)] < 0.7+n

 

[stapel]

can somebody help me to prove this inequality using the mathematical induction :

n < [n/(2n-1)+(n+1)/(2n+1)+(n+2)/(2n+3)+⋯+(3n-3)/(6n-7)+(3n-2)/(6n-5)+(3n-1)/(6n-3)] < 0.7+n

What are your thoughts? What have you tried? How far have you gotten? Where are you stuck?

Please be complete. Thank you!

 

[erblina]

What are your thoughts? What have you tried? How far have you gotten? Where are you stuck?

Please be complete. Thank you!

i tried for n=1 , n=k and tried to prove for n=k+1 but there came a point where i had to prove that one expression should be proved that is bigger than 1 to prove the left side of the inequality and smaller than 1 at the same time to prove the right side which seems impossible to do

 

[stapel]

i tried for n=1 , n=k and tried to prove for n=k+1 but there came a point where i had to prove that one expression should be proved that is bigger than 1 to prove the left side of the inequality and smaller than 1 at the same time to prove the right side which seems impossible to do

Please reply showing what you have done so far. For instance, my "for n = 1" step might look like this:

. . . . .\(\displaystyle \mbox{Let }\, n\, =\, 1.\, \mbox{ Then we have fractions with numerators}\)

. . . . .\(\displaystyle \mbox{from }\, n\, =\, 1\, \mbox{ through }\, 3n\, -\, 1\, =\, 3\, -\, 1\, =\, 2.\)

. . . . .\(\displaystyle \mbox{In other words, we need only the first two fractions:}\)

. . . . . . . .\(\displaystyle \dfrac{1}{2\, -\, 1}\, +\, \dfrac{1\, +\, 1}{2\, +\, 1}\, =\, \dfrac{1}{1}\, +\, \dfrac{2}{3}\, =\, \dfrac{5}{3}\)

. . . . .\(\displaystyle \mbox{The left-hand side inequality is:}\)

. . . . . . . .\(\displaystyle n\, =\, 1\, <\, \dfrac{5}{3}\)

. . . . .\(\displaystyle \mbox{So the left-hand inequality holds.}\)

. . . . .\(\displaystyle \mbox{The right-hand end's value is:}\)

. . . . . . . .\(\displaystyle 0.7\, +\, n\, =\, 0.7\, +\, 1\, =\, 1.7\, =\, \dfrac{17}{10}\)

. . . . .\(\displaystyle \mbox{Comparing the middle value with the right-hand end's, we get:}\)

. . . . . . . .\(\displaystyle \dfrac{5}{3}\, =\, \dfrac{50}{30}\, <\, \dfrac{51}{30}\, =\, \dfrac{17}{10}\)

. . . . .\(\displaystyle \mbox{So the right-hand inequality also holds.}\)

. . . . .\(\displaystyle \mbox{Thus, the statement holds for }\, n\, =\, 1.\)

And so forth. Please reply starting with your n = k + 1 statement, and then show your attempt at the proof. Thank you!

 

[erblina]

Please reply showing what you have done so far. For instance, my "for n = 1" step might look like this:

. . . . .\(\displaystyle \mbox{Let }\, n\, =\, 1.\, \mbox{ Then we have fractions with numerators}\)

. . . . .\(\displaystyle \mbox{from }\, n\, =\, 1\, \mbox{ through }\, 3n\, -\, 1\, =\, 3\, -\, 1\, =\, 2.\)

. . . . .\(\displaystyle \mbox{In other words, we need only the first two fractions:}\)

. . . . . . . .\(\displaystyle \dfrac{1}{2\, -\, 1}\, +\, \dfrac{1\, +\, 1}{2\, +\, 1}\, =\, \dfrac{1}{1}\, +\, \dfrac{2}{3}\, =\, \dfrac{5}{3}\)

. . . . .\(\displaystyle \mbox{The left-hand side inequality is:}\)

. . . . . . . .\(\displaystyle n\, =\, 1\, <\, \dfrac{5}{3}\)

. . . . .\(\displaystyle \mbox{So the left-hand inequality holds.}\)

. . . . .\(\displaystyle \mbox{The right-hand end's value is:}\)

. . . . . . . .\(\displaystyle 0.7\, +\, n\, =\, 0.7\, +\, 1\, =\, 1.7\, =\, \dfrac{17}{10}\)

. . . . .\(\displaystyle \mbox{Comparing the middle value with the right-hand end's, we get:}\)

. . . . . . . .\(\displaystyle \dfrac{5}{3}\, =\, \dfrac{50}{30}\, <\, \dfrac{51}{30}\, =\, \dfrac{17}{10}\)

. . . . .\(\displaystyle \mbox{So the right-hand inequality also holds.}\)

. . . . .\(\displaystyle \mbox{Thus, the statement holds for }\, n\, =\, 1.\)

And so forth. Please reply starting with your n = k + 1 statement, and then show your attempt at the proof. Thank you!

thank you very much this was a great explanation , now i'm gonna show you where my work stops :

 

[stapel]

I'm sorry, but the image is too tiny for me to be able to read.

 

[Ishuda]

can somebody help me to prove this inequality using the mathematical induction :

n < [n/(2n-1)+(n+1)/(2n+1)+(n+2)/(2n+3)+⋯+(3n-3)/(6n-7)+(3n-2)/(6n-5)+(3n-1)/(6n-3)] < 0.7+n

O.K., not doing it by induction but lets play around a bit. The sum in the middle is
\(\displaystyle s_n = \Sigma_{j = 0}^{j = 2 n - 1} \frac{n+j}{2 (n + j) -1}\)
and we can write our expression as
n < s_n < 0.7 + n
Letting N_n = n + j, we can write
N_n = 2 (n + j) - 1 + 1 - (n + j) = 2 (n + j) - 1 + 1 - N_n
or
N_n = \(\displaystyle \frac{1}{2} [ [2 (n+j) - 1] + 1 ]\)
and we can write
\(\displaystyle s_n = \frac{1}{2} \Sigma_{j = 0}^{j = 2 n - 1} \frac{2(n+j) - 1 + 1}{2 (n + j) -1} = \frac{1}{2}\Sigma_{j = 0}^{j = 2 n - 1} [ 1 + \frac{1}{2 (n + j) -1} = n + \frac{1}{2} \Sigma_{j = 0}^{j = 2 n - 1} \frac{1}{2 (n + j) -1} = n + t_n \)
So we can now write our expression as
n < n + t_n < 0.7 + n
or, since t_n is obviously greater than zero, all we have left to prove is
t_n < 0.7

Now I seem to remember something about comparing the sum of the harmonic series to the natural log function.