Mixing
Induction of $sum^n_{k=2} 1 - 1/k^2 =(1+n)/(2n)$
$begingroup$
Can someone help me with this induction please?
$$sum^n_{k=2} 1 - 1/k^2 ={1+nover 2n}.$$
sequences-and-series induction
edited Jan 18 at 18:38
peter a g
3,1601614
asked Jan 18 at 18:35
Stefan BStefan B
32
$endgroup$
|
show 1 more comment
$begingroup$
Can someone help me with this induction please?
$$sum^n_{k=2} 1 - 1/k^2 ={1+nover 2n}.$$
sequences-and-series induction
edited Jan 18 at 18:38
peter a g
3,1601614
asked Jan 18 at 18:35
Stefan BStefan B
32
$endgroup$
6
$begingroup$
What have you tried so far?
$endgroup$
– Jack Pfaffinger
Jan 18 at 18:39
1
$begingroup$
The left hand side is greater than $n-2$ and the right hand side is $frac12+frac1{2n}$ which is less than $1$ for $ngt1$. This cannot hold for many large $n$.
$endgroup$
– robjohn♦
Jan 18 at 18:44
$begingroup$
I have proved the base case for P(2). Using the hint of N. S. But i'm stack here i have replaced $sum_{i=2}^n (1 - frac{1}{k^2})$ with $frac{1 + n}{2n}$ and now i have : $frac{1 + 2}{2n} + 1 - frac{1 }{(n+1)^2}$ and i have $frac{(1+n)^3 + 2n(n+1)^2 -2n}{2n(n+1)^2} $
$endgroup$
– Stefan B
Jan 18 at 19:16
$begingroup$
Check the PS to my answer....
$endgroup$
– N. S.
Jan 18 at 23:43
$begingroup$
When n=3 the LHS is 59/36 and the RHS is 2/3, so your statement is false
$endgroup$
– jjagmath
Jan 19 at 3:48
|
show 1 more comment
$begingroup$
Can someone help me with this induction please?
$$sum^n_{k=2} 1 - 1/k^2 ={1+nover 2n}.$$
sequences-and-series induction
edited Jan 18 at 18:38
peter a g
3,1601614
asked Jan 18 at 18:35
Stefan BStefan B
32
$endgroup$
Can someone help me with this induction please?
$$sum^n_{k=2} 1 - 1/k^2 ={1+nover 2n}.$$
sequences-and-series induction
sequences-and-series induction
edited Jan 18 at 18:38
peter a g
3,1601614
asked Jan 18 at 18:35
Stefan BStefan B
32
edited Jan 18 at 18:38
peter a g
3,1601614
asked Jan 18 at 18:35
Stefan BStefan B
32
edited Jan 18 at 18:38
peter a g
3,1601614
edited Jan 18 at 18:38
peter a g
3,1601614
edited Jan 18 at 18:38
peter a g
3,1601614
3,1601614
asked Jan 18 at 18:35
Stefan BStefan B
32
asked Jan 18 at 18:35
Stefan BStefan B
32
asked Jan 18 at 18:35
Stefan BStefan B
32
32
6
$begingroup$
What have you tried so far?
$endgroup$
– Jack Pfaffinger
Jan 18 at 18:39
1
$begingroup$
The left hand side is greater than $n-2$ and the right hand side is $frac12+frac1{2n}$ which is less than $1$ for $ngt1$. This cannot hold for many large $n$.
$endgroup$
– robjohn♦
Jan 18 at 18:44
$begingroup$
I have proved the base case for P(2). Using the hint of N. S. But i'm stack here i have replaced $sum_{i=2}^n (1 - frac{1}{k^2})$ with $frac{1 + n}{2n}$ and now i have : $frac{1 + 2}{2n} + 1 - frac{1 }{(n+1)^2}$ and i have $frac{(1+n)^3 + 2n(n+1)^2 -2n}{2n(n+1)^2} $
$endgroup$
– Stefan B
Jan 18 at 19:16
$begingroup$
Check the PS to my answer....
$endgroup$
– N. S.
Jan 18 at 23:43
$begingroup$
When n=3 the LHS is 59/36 and the RHS is 2/3, so your statement is false
$endgroup$
– jjagmath
Jan 19 at 3:48
|
show 1 more comment
6
$begingroup$
What have you tried so far?
$endgroup$
– Jack Pfaffinger
Jan 18 at 18:39
1
$begingroup$
The left hand side is greater than $n-2$ and the right hand side is $frac12+frac1{2n}$ which is less than $1$ for $ngt1$. This cannot hold for many large $n$.
$endgroup$
– robjohn♦
Jan 18 at 18:44
$begingroup$
I have proved the base case for P(2). Using the hint of N. S. But i'm stack here i have replaced $sum_{i=2}^n (1 - frac{1}{k^2})$ with $frac{1 + n}{2n}$ and now i have : $frac{1 + 2}{2n} + 1 - frac{1 }{(n+1)^2}$ and i have $frac{(1+n)^3 + 2n(n+1)^2 -2n}{2n(n+1)^2} $
$endgroup$
– Stefan B
Jan 18 at 19:16
$begingroup$
Check the PS to my answer....
$endgroup$
– N. S.
Jan 18 at 23:43
$begingroup$
When n=3 the LHS is 59/36 and the RHS is 2/3, so your statement is false
$endgroup$
– jjagmath
Jan 19 at 3:48
6
6
$begingroup$
What have you tried so far?
$endgroup$
– Jack Pfaffinger
Jan 18 at 18:39
$begingroup$
What have you tried so far?
$endgroup$
– Jack Pfaffinger
Jan 18 at 18:39
1
1
$begingroup$
The left hand side is greater than $n-2$ and the right hand side is $frac12+frac1{2n}$ which is less than $1$ for $ngt1$. This cannot hold for many large $n$.
$endgroup$
– robjohn♦
Jan 18 at 18:44
$begingroup$
The left hand side is greater than $n-2$ and the right hand side is $frac12+frac1{2n}$ which is less than $1$ for $ngt1$. This cannot hold for many large $n$.
$endgroup$
– robjohn♦
Jan 18 at 18:44
$begingroup$
I have proved the base case for P(2). Using the hint of N. S. But i'm stack here i have replaced $sum_{i=2}^n (1 - frac{1}{k^2})$ with $frac{1 + n}{2n}$ and now i have : $frac{1 + 2}{2n} + 1 - frac{1 }{(n+1)^2}$ and i have $frac{(1+n)^3 + 2n(n+1)^2 -2n}{2n(n+1)^2} $
$endgroup$
– Stefan B
Jan 18 at 19:16
$begingroup$
I have proved the base case for P(2). Using the hint of N. S. But i'm stack here i have replaced $sum_{i=2}^n (1 - frac{1}{k^2})$ with $frac{1 + n}{2n}$ and now i have : $frac{1 + 2}{2n} + 1 - frac{1 }{(n+1)^2}$ and i have $frac{(1+n)^3 + 2n(n+1)^2 -2n}{2n(n+1)^2} $
$endgroup$
– Stefan B
Jan 18 at 19:16
$begingroup$
Check the PS to my answer....
$endgroup$
– N. S.
Jan 18 at 23:43
$begingroup$
Check the PS to my answer....
$endgroup$
– N. S.
Jan 18 at 23:43
$begingroup$
When n=3 the LHS is 59/36 and the RHS is 2/3, so your statement is false
$endgroup$
– jjagmath
Jan 19 at 3:48
$begingroup$
When n=3 the LHS is 59/36 and the RHS is 2/3, so your statement is false
$endgroup$
– jjagmath
Jan 19 at 3:48
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
Hint
$$sum^{n+1}_{k=2} 1 -frac{1}{k^2}=left(sum^{n}_{k=2} 1 -frac{1}{k^2} right) +1-frac{1}{(n+1)^2}$$
Once you fix your problem (note that the sequence on the LHS is increasing approximately ar the rate of $S_n sim n$, while the RHS is tending to a constant, you probably copied the problem wrong) just prove $P(2)$ and use the above formula to prove the inductive step.
P.S. Are you sure the exercise is NOT
$$prod^{n}_{k=2}( 1 -frac{1}{k^2})=frac{n+1}{2n}$$
if it is, note
$$prod^{n+1}_{k=2} (1 -frac{1}{k^2})=(1 -frac{1}{(n+1)^2})prod^{n}_{k=2} (1 -frac{1}{k^2})=(frac{n(n+2)}{(n+1)^2})prod^{n}_{k=2} (1 -frac{1}{k^2})
$$
edited Jan 19 at 1:49
peter a g
3,1601614
answered Jan 18 at 18:41
N. S.N. S.
104k7114209
$endgroup$
$begingroup$
Thank you for the hint! I have proved the formula for P(2). Now i'm stack here i have replaced $sum_{i=2}^n (1 - frac{1}{k^2})$ with $frac{1 + n}{2n}$ and now i have : $frac{1 + 2}{2n} + 1 - frac{1 }{(n+1)^2}$ and i have $frac{(1+n)^3 + 2n(n+1)^2 -2n}{2n(n+1)^2} $
$endgroup$
– Stefan B
Jan 18 at 19:11
$begingroup$
@StefanB Again there is a mistake in the problem...The LHS is divergent, the RHS in your formula is convergent...Are you sure the formula is a sum, not a product? I think with product it is correct.
$endgroup$
– N. S.
Jan 18 at 23:35
$begingroup$
@N.S. I hope you don't mind: I corrected mathematical typos. See the revision history.
$endgroup$
– peter a g
Jan 19 at 1:51
$begingroup$
Thank you again yes I think that I have copied the problem wrong. With the product it makes sense. Thank you again
$endgroup$
– Stefan B
Jan 19 at 10:56
add a comment |
$begingroup$
Assuming that this is a summation and not a product, if you enjoy generalized harmonic numbers,
$$S_n=sum^n_{k=2}left( 1 - frac 1 {k^2} right)=n-H_n^{(2)}$$ For large values of $n$, this would give
$$S_n=-frac{pi ^2}{6}+n+frac{1}{n}-frac{1}{2
n^2}+Oleft(frac{1}{n^3}right)$$ which has little to do with $frac{n+1}{2n}$.
It must be the product as already said in comments and answers.
answered Jan 19 at 6:54
Claude LeiboviciClaude Leibovici
123k1157134
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint
$$sum^{n+1}_{k=2} 1 -frac{1}{k^2}=left(sum^{n}_{k=2} 1 -frac{1}{k^2} right) +1-frac{1}{(n+1)^2}$$
Once you fix your problem (note that the sequence on the LHS is increasing approximately ar the rate of $S_n sim n$, while the RHS is tending to a constant, you probably copied the problem wrong) just prove $P(2)$ and use the above formula to prove the inductive step.
P.S. Are you sure the exercise is NOT
$$prod^{n}_{k=2}( 1 -frac{1}{k^2})=frac{n+1}{2n}$$
if it is, note
$$prod^{n+1}_{k=2} (1 -frac{1}{k^2})=(1 -frac{1}{(n+1)^2})prod^{n}_{k=2} (1 -frac{1}{k^2})=(frac{n(n+2)}{(n+1)^2})prod^{n}_{k=2} (1 -frac{1}{k^2})
$$
edited Jan 19 at 1:49
peter a g
3,1601614
answered Jan 18 at 18:41
N. S.N. S.
104k7114209
$endgroup$
$begingroup$
Thank you for the hint! I have proved the formula for P(2). Now i'm stack here i have replaced $sum_{i=2}^n (1 - frac{1}{k^2})$ with $frac{1 + n}{2n}$ and now i have : $frac{1 + 2}{2n} + 1 - frac{1 }{(n+1)^2}$ and i have $frac{(1+n)^3 + 2n(n+1)^2 -2n}{2n(n+1)^2} $
$endgroup$
– Stefan B
Jan 18 at 19:11
$begingroup$
@StefanB Again there is a mistake in the problem...The LHS is divergent, the RHS in your formula is convergent...Are you sure the formula is a sum, not a product? I think with product it is correct.
$endgroup$
– N. S.
Jan 18 at 23:35
$begingroup$
@N.S. I hope you don't mind: I corrected mathematical typos. See the revision history.
$endgroup$
– peter a g
Jan 19 at 1:51
$begingroup$
Thank you again yes I think that I have copied the problem wrong. With the product it makes sense. Thank you again
$endgroup$
– Stefan B
Jan 19 at 10:56
add a comment |
$begingroup$
Hint
$$sum^{n+1}_{k=2} 1 -frac{1}{k^2}=left(sum^{n}_{k=2} 1 -frac{1}{k^2} right) +1-frac{1}{(n+1)^2}$$
Once you fix your problem (note that the sequence on the LHS is increasing approximately ar the rate of $S_n sim n$, while the RHS is tending to a constant, you probably copied the problem wrong) just prove $P(2)$ and use the above formula to prove the inductive step.
P.S. Are you sure the exercise is NOT
$$prod^{n}_{k=2}( 1 -frac{1}{k^2})=frac{n+1}{2n}$$
if it is, note
$$prod^{n+1}_{k=2} (1 -frac{1}{k^2})=(1 -frac{1}{(n+1)^2})prod^{n}_{k=2} (1 -frac{1}{k^2})=(frac{n(n+2)}{(n+1)^2})prod^{n}_{k=2} (1 -frac{1}{k^2})
$$
edited Jan 19 at 1:49
peter a g
3,1601614
answered Jan 18 at 18:41
N. S.N. S.
104k7114209
$endgroup$
$begingroup$
Thank you for the hint! I have proved the formula for P(2). Now i'm stack here i have replaced $sum_{i=2}^n (1 - frac{1}{k^2})$ with $frac{1 + n}{2n}$ and now i have : $frac{1 + 2}{2n} + 1 - frac{1 }{(n+1)^2}$ and i have $frac{(1+n)^3 + 2n(n+1)^2 -2n}{2n(n+1)^2} $
$endgroup$
– Stefan B
Jan 18 at 19:11
$begingroup$
@StefanB Again there is a mistake in the problem...The LHS is divergent, the RHS in your formula is convergent...Are you sure the formula is a sum, not a product? I think with product it is correct.
$endgroup$
– N. S.
Jan 18 at 23:35
$begingroup$
@N.S. I hope you don't mind: I corrected mathematical typos. See the revision history.
$endgroup$
– peter a g
Jan 19 at 1:51
$begingroup$
Thank you again yes I think that I have copied the problem wrong. With the product it makes sense. Thank you again
$endgroup$
– Stefan B
Jan 19 at 10:56
add a comment |
$begingroup$
Hint
$$sum^{n+1}_{k=2} 1 -frac{1}{k^2}=left(sum^{n}_{k=2} 1 -frac{1}{k^2} right) +1-frac{1}{(n+1)^2}$$
Once you fix your problem (note that the sequence on the LHS is increasing approximately ar the rate of $S_n sim n$, while the RHS is tending to a constant, you probably copied the problem wrong) just prove $P(2)$ and use the above formula to prove the inductive step.
P.S. Are you sure the exercise is NOT
$$prod^{n}_{k=2}( 1 -frac{1}{k^2})=frac{n+1}{2n}$$
if it is, note
$$prod^{n+1}_{k=2} (1 -frac{1}{k^2})=(1 -frac{1}{(n+1)^2})prod^{n}_{k=2} (1 -frac{1}{k^2})=(frac{n(n+2)}{(n+1)^2})prod^{n}_{k=2} (1 -frac{1}{k^2})
$$
edited Jan 19 at 1:49
peter a g
3,1601614
answered Jan 18 at 18:41
N. S.N. S.
104k7114209
$endgroup$
Hint
$$sum^{n+1}_{k=2} 1 -frac{1}{k^2}=left(sum^{n}_{k=2} 1 -frac{1}{k^2} right) +1-frac{1}{(n+1)^2}$$
Once you fix your problem (note that the sequence on the LHS is increasing approximately ar the rate of $S_n sim n$, while the RHS is tending to a constant, you probably copied the problem wrong) just prove $P(2)$ and use the above formula to prove the inductive step.
P.S. Are you sure the exercise is NOT
$$prod^{n}_{k=2}( 1 -frac{1}{k^2})=frac{n+1}{2n}$$
if it is, note
$$prod^{n+1}_{k=2} (1 -frac{1}{k^2})=(1 -frac{1}{(n+1)^2})prod^{n}_{k=2} (1 -frac{1}{k^2})=(frac{n(n+2)}{(n+1)^2})prod^{n}_{k=2} (1 -frac{1}{k^2})
$$
edited Jan 19 at 1:49
peter a g
3,1601614
answered Jan 18 at 18:41
N. S.N. S.
104k7114209
edited Jan 19 at 1:49
peter a g
3,1601614
edited Jan 19 at 1:49
peter a g
3,1601614
edited Jan 19 at 1:49
peter a g
3,1601614
3,1601614
answered Jan 18 at 18:41
N. S.N. S.
104k7114209
answered Jan 18 at 18:41
N. S.N. S.
104k7114209
answered Jan 18 at 18:41
N. S.N. S.
104k7114209
104k7114209
$begingroup$
Thank you for the hint! I have proved the formula for P(2). Now i'm stack here i have replaced $sum_{i=2}^n (1 - frac{1}{k^2})$ with $frac{1 + n}{2n}$ and now i have : $frac{1 + 2}{2n} + 1 - frac{1 }{(n+1)^2}$ and i have $frac{(1+n)^3 + 2n(n+1)^2 -2n}{2n(n+1)^2} $
$endgroup$
– Stefan B
Jan 18 at 19:11
$begingroup$
@StefanB Again there is a mistake in the problem...The LHS is divergent, the RHS in your formula is convergent...Are you sure the formula is a sum, not a product? I think with product it is correct.
$endgroup$
– N. S.
Jan 18 at 23:35
$begingroup$
@N.S. I hope you don't mind: I corrected mathematical typos. See the revision history.
$endgroup$
– peter a g
Jan 19 at 1:51
$begingroup$
Thank you again yes I think that I have copied the problem wrong. With the product it makes sense. Thank you again
$endgroup$
– Stefan B
Jan 19 at 10:56
add a comment |
$begingroup$
Thank you for the hint! I have proved the formula for P(2). Now i'm stack here i have replaced $sum_{i=2}^n (1 - frac{1}{k^2})$ with $frac{1 + n}{2n}$ and now i have : $frac{1 + 2}{2n} + 1 - frac{1 }{(n+1)^2}$ and i have $frac{(1+n)^3 + 2n(n+1)^2 -2n}{2n(n+1)^2} $
$endgroup$
– Stefan B
Jan 18 at 19:11
$begingroup$
@StefanB Again there is a mistake in the problem...The LHS is divergent, the RHS in your formula is convergent...Are you sure the formula is a sum, not a product? I think with product it is correct.
$endgroup$
– N. S.
Jan 18 at 23:35
$begingroup$
@N.S. I hope you don't mind: I corrected mathematical typos. See the revision history.
$endgroup$
– peter a g
Jan 19 at 1:51
$begingroup$
Thank you again yes I think that I have copied the problem wrong. With the product it makes sense. Thank you again
$endgroup$
– Stefan B
Jan 19 at 10:56
$begingroup$
Thank you for the hint! I have proved the formula for P(2). Now i'm stack here i have replaced $sum_{i=2}^n (1 - frac{1}{k^2})$ with $frac{1 + n}{2n}$ and now i have : $frac{1 + 2}{2n} + 1 - frac{1 }{(n+1)^2}$ and i have $frac{(1+n)^3 + 2n(n+1)^2 -2n}{2n(n+1)^2} $
$endgroup$
– Stefan B
Jan 18 at 19:11
$begingroup$
Thank you for the hint! I have proved the formula for P(2). Now i'm stack here i have replaced $sum_{i=2}^n (1 - frac{1}{k^2})$ with $frac{1 + n}{2n}$ and now i have : $frac{1 + 2}{2n} + 1 - frac{1 }{(n+1)^2}$ and i have $frac{(1+n)^3 + 2n(n+1)^2 -2n}{2n(n+1)^2} $
$endgroup$
– Stefan B
Jan 18 at 19:11
$begingroup$
@StefanB Again there is a mistake in the problem...The LHS is divergent, the RHS in your formula is convergent...Are you sure the formula is a sum, not a product? I think with product it is correct.
$endgroup$
– N. S.
Jan 18 at 23:35
$begingroup$
@StefanB Again there is a mistake in the problem...The LHS is divergent, the RHS in your formula is convergent...Are you sure the formula is a sum, not a product? I think with product it is correct.
$endgroup$
– N. S.
Jan 18 at 23:35
$begingroup$
@N.S. I hope you don't mind: I corrected mathematical typos. See the revision history.
$endgroup$
– peter a g
Jan 19 at 1:51
$begingroup$
@N.S. I hope you don't mind: I corrected mathematical typos. See the revision history.
$endgroup$
– peter a g
Jan 19 at 1:51
$begingroup$
Thank you again yes I think that I have copied the problem wrong. With the product it makes sense. Thank you again
$endgroup$
– Stefan B
Jan 19 at 10:56
$begingroup$
Thank you again yes I think that I have copied the problem wrong. With the product it makes sense. Thank you again
$endgroup$
– Stefan B
Jan 19 at 10:56
add a comment |
$begingroup$
Assuming that this is a summation and not a product, if you enjoy generalized harmonic numbers,
$$S_n=sum^n_{k=2}left( 1 - frac 1 {k^2} right)=n-H_n^{(2)}$$ For large values of $n$, this would give
$$S_n=-frac{pi ^2}{6}+n+frac{1}{n}-frac{1}{2
n^2}+Oleft(frac{1}{n^3}right)$$ which has little to do with $frac{n+1}{2n}$.
It must be the product as already said in comments and answers.
answered Jan 19 at 6:54
Claude LeiboviciClaude Leibovici
123k1157134
$endgroup$
add a comment |
$begingroup$
Assuming that this is a summation and not a product, if you enjoy generalized harmonic numbers,
$$S_n=sum^n_{k=2}left( 1 - frac 1 {k^2} right)=n-H_n^{(2)}$$ For large values of $n$, this would give
$$S_n=-frac{pi ^2}{6}+n+frac{1}{n}-frac{1}{2
n^2}+Oleft(frac{1}{n^3}right)$$ which has little to do with $frac{n+1}{2n}$.
It must be the product as already said in comments and answers.
answered Jan 19 at 6:54
Claude LeiboviciClaude Leibovici
123k1157134
$endgroup$
add a comment |
$begingroup$
Assuming that this is a summation and not a product, if you enjoy generalized harmonic numbers,
$$S_n=sum^n_{k=2}left( 1 - frac 1 {k^2} right)=n-H_n^{(2)}$$ For large values of $n$, this would give
$$S_n=-frac{pi ^2}{6}+n+frac{1}{n}-frac{1}{2
n^2}+Oleft(frac{1}{n^3}right)$$ which has little to do with $frac{n+1}{2n}$.
It must be the product as already said in comments and answers.
answered Jan 19 at 6:54
Claude LeiboviciClaude Leibovici
123k1157134
$endgroup$
Assuming that this is a summation and not a product, if you enjoy generalized harmonic numbers,
$$S_n=sum^n_{k=2}left( 1 - frac 1 {k^2} right)=n-H_n^{(2)}$$ For large values of $n$, this would give
$$S_n=-frac{pi ^2}{6}+n+frac{1}{n}-frac{1}{2
n^2}+Oleft(frac{1}{n^3}right)$$ which has little to do with $frac{n+1}{2n}$.
It must be the product as already said in comments and answers.
answered Jan 19 at 6:54
Claude LeiboviciClaude Leibovici
123k1157134
answered Jan 19 at 6:54
Claude LeiboviciClaude Leibovici
123k1157134
answered Jan 19 at 6:54
Claude LeiboviciClaude Leibovici
123k1157134
answered Jan 19 at 6:54
Claude LeiboviciClaude Leibovici
123k1157134
123k1157134
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6
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What have you tried so far?
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– Jack Pfaffinger
Jan 18 at 18:39
1
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The left hand side is greater than $n-2$ and the right hand side is $frac12+frac1{2n}$ which is less than $1$ for $ngt1$. This cannot hold for many large $n$.
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– robjohn♦
Jan 18 at 18:44
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I have proved the base case for P(2). Using the hint of N. S. But i'm stack here i have replaced $sum_{i=2}^n (1 - frac{1}{k^2})$ with $frac{1 + n}{2n}$ and now i have : $frac{1 + 2}{2n} + 1 - frac{1 }{(n+1)^2}$ and i have $frac{(1+n)^3 + 2n(n+1)^2 -2n}{2n(n+1)^2} $
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– Stefan B
Jan 18 at 19:16
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Check the PS to my answer....
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– N. S.
Jan 18 at 23:43
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When n=3 the LHS is 59/36 and the RHS is 2/3, so your statement is false
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– jjagmath
Jan 19 at 3:48